Sequences And Series Of Real Numbers Class 10th Mathematics Tamilnadu Board Solution

Class 10^th Mathematics Tamilnadu Board Solution

Exercise 2.1

1. Write the first three terms of the following sequences whose nth terms are given…
2. Find the indicated terms in each of the sequences whose nth terms are given by…
3. Find the 18th and 25th terms of the sequence defined by a_n = ll n (n+3) , & n…
4. Find the 13th and 16th terms of the sequence defined by b_n = cl n^2 , & n inn n…
5. Find the first five terms of the sequence given by a1 = 2, a2 = 3 + a1and an =…
6. Find the first six terms of the sequence given by a1 = a2 = a3 = 1 and an = an-1…

Exercise 2.2

1. The first term of an A.P. is 6 and the common difference is 5. Find the A.P. and…
2. Find the common difference and 15th term of the A.P. 125, 120, 115, 110 ……
3. Which term of the arithmetic sequence 24 , 23 1/4 , 22 1/2 , 21 3/4 , l is 3?…
4. Find the 12th term of the A.P. root 2 , 3 root 2 , 5 root 2 , .
5. Find the 17th term of the A.P. 4, 9, 14 …
6. How many terms are there in the following Arithmetic Progressions? (i) -1 , -…
7. If 9th term of an A.P. is zero, prove that its 29th term is double (twice) the…
8. The 10th and 18th terms of an A.P are 41 and 73 respectively. Find the 27th…
9. Find n so that the nth terms of the following two A.P.’s are the same. 1, 7, 13,…
10. How many two digit numbers are divisible by 13?
11. A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the…
12. A man has saved ₹640 during the first month, ₹720 in the second month and ₹800…
13. The sum of three consecutive terms in an A.P. is 6 and their product is -120.…
14. Find the three consecutive terms in an A. P. whose sum is 18 and the sum of…
15. If m times the mth term of an A.P. is equal to n times its nth term, then show…
16. A person has deposited ₹25,000 in an investment which yields 14% simple…
17. If a, b, c are in A.P. then prove that (a - c)^2 = 4 (b^2 - ac).
18. If a, b, c are in A.P. then prove that 1/bc , 1/ca , 1/ab are also in A.P.…
19. If a^2 , b^2 , c^2 are in A.P. then show that 1/b+c , 1/c+a , 1/a+b are also in…
20. If ax = by = cz, x ≠ 0, y ≠ 0, z ≠ 0 and b^2 = ac, then show that 1/x , 1/y ,…

Exercise 2.3

1. 0.12, 0.24, 0.48,…. Find out which of the following sequences are geometric…
2. 0.004, 0.02, 0.1,…. Find out which of the following sequences are geometric…
3. 1/2 , 1/3 , 2/9 , 4/27 ….. . Find out which of the following sequences are…
4. 12 , 1 , 1/12 , l Find out which of the following sequences are geometric…
5. root 2 , 1/root 2 , 1/2 root 2 , Find out which of the following sequences are…
6. 4 ,-2 ,-1 , - 1/2 , l Find out which of the following sequences are geometric…
7. Find the 10th term and common ratio of the geometric sequence 1/4 , - 1/2 , 1…
8. If the 4th and 7th terms of a G.P. are 54 and 1458 respectively, find the G.P.…
9. In a geometric sequence, the first term is 1/3 and the sixth term is 1/729 ,…
10. Which term of the geometric sequence, (i) 5, 2, 4/5 , 8/25 … , is 128/15625 ?…
11. If the geometric sequences 162, 54, 18,…. and 2/81 , 2/27 , 2/9 ,…have their nth…
12. The fifth term of a G.P. is 1875. If the first term is 3, find the common ratio.…
13. The sum of three terms of a geometric sequence is 39/10 and their product is 1.…
14. If the product of three consecutive terms in G.P. is 216 and sum of their…
15. Find the first three consecutive terms in G.P. whose sum is 7 and the sum of…
16. The sum of the first three terms of a G.P. is 13 and sum of their squares is…
17. If ₹1000 is deposited in a bank which pays annual interest at the rate of 5%…
18. A company purchases an office copier machine for ₹50,000. It is estimated that…
19. If a, b, c, d are in a geometric sequence, then show that (a - b + c) (b + c +…
20. If a, b, c, d are in a G.P., then prove that a + b, b + c, c + d, are also in…

Exercise 2.4

1. Find the sum of the first (i) 75 positive integers (ii) 125 natural numbers.…
2. Find the sum of the first 30 terms of an A.P. whose nth term is 3 + 2n.…
3. Find the sum of each arithmetic series (i) 38 + 35 + 32 + … + 2. (ii) 6+5 1/4 +…
4. Find the Snfor the following arithmetic series described. (i) a = 5, n = 30, l =…
5. Find the sum of the first 40 terms of the series 1^2 - 2^2 + 3^2 - 4^2 + … .…
6. In an arithmetic series, the sum of first 11 terms is 44 and that of the next 11…
7. In the arithmetic sequence 60, 56, 52, 48,…, starting from the first term, how…
8. Find the sum of all 3 digit natural numbers, which are divisible by 9.…
9. Find the sum of first 20 terms of the arithmetic series in which 3rd term is 7…
10. Find the sum of all natural numbers between 300 and 500 which are divisible by…
11. Solve: 1 + 6 + 11 + 16 + ….. + x = 148.
12. Find the sum of all numbers between 100 and 200 which are not divisible by 5.…
13. A construction company will be penalised each day for delay in construction of…
14. A sum of ₹1000 is deposited every year at 8% simple interest. Calculate the…
15. The sum of first n terms of a certain series is given as 3n^2 - 2n. Show that…
16. If a clock strikes once at 1 o’clock, twice at 2 o’clock and so on, how many…
17. Show that the sum of an arithmetic series whose first term is a, second term b…
18. If there are (2n + 1) terms in an arithmetic series, then prove that the ratio…
19. The ratio of the sums of first m and first n terms of an arithmetic series is…
20. A gardener plans to construct a trapezoidal shaped structure in his garden. The…

Exercise 2.5

1. Find the sum of the first 20 terms of the geometric series 5/2 + 5/6 + 5/18 + l…
2. Find the sum of the first 27 terms of the geometric series 1/9 + 1/27 + 1/81 + l…
3. Find Snfor each of the geometric series described below. (i) a = 3, t8 = 384, n…
4. Find the sum of the following finite series (i) 1 + 0.1 + 0.01 + 0.001 + … +…
5. How many consecutive terms starting from the first term of the series (i) 3 + 9…
6. The second term of a geometric series is 3 and the common ratio is 4/5 . Find…
7. A geometric series consists of four terms and has a positive common ratio. The…
8. Find the sum of first n terms of the series (i) 7 + 77 + 777 + … . (ii) 0.4 +…
9. Suppose that five people are ill during the first week of an epidemic and each…
10. A gardener wanted to reward a boy for his good deeds by giving some mangoes. He…
11. A geometric series consists of even number of terms. The sum of all terms is 3…
12. If S1,S2and S3 are the sum of first n, 2n and 3n terms of a geometric series…

Exercise 2.6

1. 1 + 2 + 3 + …. + 45 Find the sum of the following series.
2. 16^2 + 17^2 + 18 + … + 25^2 Find the sum of the following series.…
3. 2 + 4 + 6 + … + 100 Find the sum of the following series.
4. 7 + 14 + 21…. + 490 Find the sum of the following series.
5. 5^2 + 7^2 + 9^2 + … + 39^2 Find the sum of the following series.
6. 16^3 + 17^3 + … + 35^3 Find the sum of the following series.
7. 1^3 + 2^3 + 3^3 + … + k^3 = 6084 Find the value of k if
8. 1^3 + 2^3 + 3^3 + … + k^3 = 2025 Find the value of k if
9. If 1 + 2 + 3 + … + p = 171, then find 1^3 + 2^3 + 3^3 + ... + p^3 .…
10. If 1^3 + 2^3 + 3^3 + … + k^3 = 8281, then find 1 + 2 + 3 + … + k.…
11. Find the total area of 12 squares whose sides are 12 cm, 13cm, g, 23cm.…
12. Find the total volume of 15 cubes whose edges are 16 cm, 17 cm, 18 cm, …., 30 cm…

Exercise 2.7

1. Which one of the following is not true?A. A sequence is a real valued function…
2. The 8th term of the sequence 1, 1, 2, 3, 5, 8, g isA. 25 B. 24 C. 23 D. 21…
3. The next term of 1/20 in the sequence 1/2 1/6 , 1/12 , 1/20 , l isA. 1/24 B.…
4. If a, b, c, l, m are in A.P, then the value of a - 4b + 6c - 4l + m isA. 1 B. 2…
5. If a, b, c are in A.P. then a-b/b-c is equal toA. a/b B. b/c C. a/c D. 1…
6. If the nth term of a sequence is 100 n + 10, then the sequence isA. an A.P. B. a…
7. If a1, a2 , a3,...are in A.P. such that a_4/a_7 = 3/2 , then the 13th term of…
8. If the sequence a1, a2, a3,… is in A.P. , then the sequence a5, a10, a15,…isA. a…
9. If k + 2, 4k-6, 3k-2 are the three consecutive terms of an A.P, then the value…
10. If a, b, c, l, m. n are in A.P., then 3a + 7, 3b + 7, 3c + 7, 3l + 7, 3m + 7,…
11. If the third term of a G.P is 2, then the product of first 5 terms isA. 5^2 B.…
12. If a, b, c are in G.P, then is equal to a-b/b-c is equal toA. a/b B. b/a C. a/c…
13. If x, 2x + 2, 3x + 3 are in G.P, then 5x, 10x + 10, 15x + 15 formA. an A.P. B.…
14. The sequence -3, -3, -3,… isA. an A.P. only B. a G.P. only C. neither A.P. nor…
15. If the product of the first four consecutive terms of a G.P is 256 and if the…
16. In a G.P, t2 = 3/5 and t3 = 1/5 . Then the common ratio isA. 1/5 B. 1/3 C. 1 D.…
17. If x not equal 0, then 1 + sec x + sec^2 x + sec^3 x + sec^4 x + sec^5 x is…
18. If the nthterm of an A.P. is tn = 3-5n, then the sum of the first n terms isA.…
19. The common ratio of the G.P. am-n, am, am + nisA. am B. a-m C. an D. a -n…
20. If 1 + 2 + 3 + . . . + n = k then 1^3 + 2^3 + … + n^3 is equal toA. k2 B. k^3…

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Exercise 2.1

Question 1.

Write the first three terms of the following sequences whose n^th terms are given by

(i)  (ii) c_n = (–1)ⁿ3ⁿ⁺² (iii) 

Answer:

(i) Here, a_n = 

For n = 1, a₁ =  =  = 

For n = 2, a₂ =  =  =  = 0

For n = 3, a₃ =  =  =  = 1

Hence, the first three terms of the sequence are, 0 and 1.

(ii) Here, c_n = (-1)ⁿ 3^{n + 2}

For n = 1, c₁ = (-1)¹ 3^{1 + 2} = (-1) 3³ = (-1) (27) = -27

For n = 2, c₂ = (-1)² 3^{2 + 2} = (1) 3⁴ = (1) (81) = 81

For n = 3, c₃ = (-1)³ 3^{3 + 2} = (-1) 3⁵ = (-1) (243) = -243

Hence, the first three terms of the sequence are -27, 81 and -243.

(iii) Here, z_n = 

For n = 1, z₁ =  =  = 

For n = 2, z₂ =  =  =  = 2

For n = 3, z₃ =  =  = 

Hence, the first three terms of the sequence are, 2 and.

Question 2.

Find the indicated terms in each of the sequences whose n^th terms are given by

(i)  (ii) a_n = (-1)ⁿ 2^{n + 3} (n + 1); a₅, a₈

(iii) a_n = 2n² – 3n + 1; a₅, a₇ (iv) a_n = (-1)ⁿ (1 – n + n²); a₅, a₈

Answer:

(i) Here, a_n = 

For n = 7, a₇ =  =  = 

For n = 9, a₉ =  =  = 

(ii) Here, a_n = (-1)ⁿ 2^{n + 3} (n + 1)

For n = 5, a₅ = (-1)⁵ 2^{5 + 3} (5 + 1) = (-1) 2⁸ (6) = -6 (256) = -1536

For n = 8, a₈ = (-1)⁸ 2^{8 + 3} (8 + 1) = (1) 2¹¹ (9) = 9 (2048)

= 18432

(iii) Here, a_n = 2n² – 3n + 1

For n = 5, a₅ = 2(5)² – 3(5) + 1 = 2(25) – 15 + 1 = 50 – 15 + 1

= 36

For n = 7, a₇ = 2(7)² – 3(7) + 1 = 2(49) – 15 + 1 = 98 – 21 + 1

= 78

(iv) Here, a_n = (-1)ⁿ (1 – n + n²)

For n = 5, a₅ = (-1)⁵ (1 – 5 + 5²) = (-1) (1 – 5 + 25) = (-1) (21)

= -21

For n = 8, a₅ = (-1)⁸ (1 – 8 + 8²) = (1) (1 – 8 + 64) = (1) (57)

= 57

Question 3.

Find the 18th and 25th terms of the sequence defined by

Answer:

For n = 18, n is even.

So, a₁₈ = 18(18 + 3) = 18(21) = 378

For n = 25, n is odd.

So, a₂₅ =  =  =  = 

Question 4.

Find the 13th and 16th terms of the sequence defined by

Answer:

For n = 13, n is odd.

So, b_n = 13(13 + 2) = 13(15) = 195

For n = 16, n is even.

So, b_n = 16² = 256

Question 5.

Find the first five terms of the sequence given by a₁ = 2, a₂ = 3 + a₁and a_n = 2a_n-1 + 5 for n>2.

Answer:

Given that a₁ = 2, a₂ = 3 + a₁ and a_n = 2a_n-1 + 5 for n > 2.

Now, a₁ = 2

⇒ a₂ = 3 + a₁ = 3 + 2 = 5

⇒ a₃ = 2a₂ + 5 = 2(5) + 5 = 10 + 5 = 15

⇒ a₄ = 2a₃ + 5 = 2(15) + 5 = 30 + 5 = 35

⇒ a₅ = 2a₄ + 5 = 2(35) + 5 = 70 + 5 = 75

∴ The required terms of sequence are 2, 5, 15, 35 and 75.

Question 6.

Find the first six terms of the sequence given by

a₁ = a₂ = a₃ = 1 and a_n = a_n-1 + a_n-2 for n >3.

Answer:

Given that a₁ = a₂ = a₃ = 1 and a_n = a_n-1 + a_n-2 for n > 3.

Now, a₁ = 1

⇒ a₂ = 1

⇒ a₃ = 1

⇒ a₄ = a₃ + a₂ = 1 + 1 = 2

⇒ a₅ = a₄ + a₃ = 2 + 1 = 3

⇒ a₆ = a₅ + a₄ = 3 + 2 = 5

∴ The required terms of the sequence are 1, 1, 1, 2, 3 and 5.

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Exercise 2.2

Question 1.

The first term of an A.P. is 6 and the common difference is 5. Find the A.P. and its general term.

Answer:

First term, a = 6; Common difference, d = 5

We know that the AP is in the form of a, a + d, a + 2d, a + 3d … a + (n – 1) d, a + nd …

⇒ AP = 6, 6 + 5, 6 + 2(5), 6 + 3(5) …

= 6, 11, 6 + 10, 6 + 15 …

= 6, 11, 16, 21 …

∴ The required A.P. is 6, 11, 16, 21 …

We know that the general form, t_n = a + (n – 1) d.

⇒ t_n = 6 + (n – 1) 5

= 6 + 5n – 5

= 5n + 1

∴ The required general term, t_n = 5n + 1

Question 2.

Find the common difference and 15th term of the A.P. 125, 120, 115, 110 …

Answer:

The given A.P. is 125, 120, 115, 110 …

We know that common difference, d = a₁ – a.

⇒ d = 120 – 125

= -5

∴ Common difference = -5

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₁₅ = 125 + (15 – 1) (-5)

= 125 + 14 (-5)

= 125 – 70

= 55

∴ 15^th term of A.P. = 55

Question 3.

Which term of the arithmetic sequence  is 3?

Answer:

Here, first term, a = 24

We know that common difference, d = a₁ – a.

We know that nth term of A.P. , t_n = a + (n – 1) d.

To find the nth term, here t_n = 3

⇒ 3 = 24 + (n – 1) (-3/4)

⇒ 3 – 24 = (n – 1) (-3/4)

⇒ -21 = (n – 1) (-3/4)

⇒ -21 ×  = n – 1

⇒ 28 = n – 1

⇒ n = 28 + 1 = 29

∴ t₂₉ of the given A.P. is 3.

Question 4.

Find the 12th term of the A.P. 

Answer:

Given A.P. √2, 3√2, 5√2 …

Here, first term, a = √2

We know that common difference, d = a₁ – a.

∴ d = 3√2 - √2 = 2√2

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₁₂ = √2 + (12 – 1) (2√2)

= √2 + 11 (2√2)

= √2 + 22 √2

= 23√2

∴ 12^th term of A.P. i.e. t₁₂ = 23√2

Question 5.

Find the 17th term of the A.P. 4, 9, 14 …

Answer:

Given A.P. 4, 9, 14 …

Here, first term, a = 4

We know that common difference, d = a₁ – a.

∴ d = 9 - 4 = 5

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₁₇ = 4 + (17 – 1) (5)

= 4 + 16 (5)

= 4 + 80

= 84

∴ 17^th term of A.P. i.e. t₁₇ = 84

Question 6.

How many terms are there in the following Arithmetic Progressions?

(i)  (ii) 7, 13, 19, …, 205.

Answer:

(i) Here, first term, a = -1

Last term, l = 

We know that common difference, d = a₁ – a.

∴ d =  – (-1) =  + 1 =  = 

We know that Number of terms, n = () + 1.

⇒ n =  + 1

=  + 1

=  + 1

= (13 × 2) + 1

= 26 + 1

= 27

∴ There are 27 terms in the given A.P.

(ii) Here, first term, a = 7

Last term, l = 205

We know that common difference, d = a₁ – a.

∴ d = 13 - 7 = 6

We know that Number of terms, n = () + 1.

⇒ n =  + 1

=  + 1

= 33 + 1

= 34

∴ There are 34 terms in the given A.P.

Question 7.

If 9th term of an A.P. is zero, prove that its 29th term is double (twice) the 19th term.

Answer:

Here, t₉ = 0

We have to prove that t₂₉ = 2t₁₉

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₉ = a + (9 – 1) d = a + 8d

But t₉ = 0

⇒ a + 8d = 0

∴ a = -8d … (i)

Now, t₂₉ = a + (29 – 1) d

= a + 28d

= -8d + 28d [From (i)]

= 20d

∴ t₂₉ = 20d … (1)

Now, t₁₉ = a + (19 – 1) d

= a + 18d

= -8d + 18d [From (i)]

= 10d

∴ t₁₉ = 10d … (2)

Equating (1) and (2),

⇒ 20d = 10d

⇒ 20d = 10d (2)

∴ t₂₉ = 2(t₁₉)

Hence proved.

Question 8.

The 10th and 18th terms of an A.P are 41 and 73 respectively. Find the 27th term.

Answer:

Here, t₁₀ = 41 and t₁₈ = 73

We know that nth term of A.P. , t_n = a + (n – 1) d.

First, t₁₀ = a + (10 – 1) d = 41

⇒ a + 9d = 41 … (1)

Then, t₁₈ = a + (18 – 1) d = 73

⇒ a + 17d = 73 … (2)

From (1) and (2),

a + 9d = 41

a + 17d = 73

(-) (-) (-)

-8d = -32

⇒ d = 32/8 = 4

Substituting d = 4 in (1),

⇒ a + 9(4) = 41

⇒ a = 41 – 36 = 5

Now, t₂₇ = 5 + (27 – 1) (4)

= 5 + 26(4)

= 5 + 104

= 109

∴ The 27^th term i.e. t₂₇ = 109

Question 9.

Find n so that the nth terms of the following two A.P.’s are the same.

1, 7, 13, 19,… and 100, 95, 90,… .

Answer:

For first A.P. , first term, a = 1; common difference, d = 7 – 1 = 6

For second A.P, first term, a = 100; common difference , d = 95 – 100 = -5

We know that nth term of A.P. , t_n = a + (n – 1) d.

First A.P:

⇒ t_n1 = 1 + (n – 1) (6)

= 1 + 6n – 6

∴ t_n1 = 6n – 5

Second A.P:

⇒ t_n2 = 100 + (n – 1) (-5)

= 100 – 5n + 5

∴ t_n2 = 105 – 5n

Given that nth terms of the two A.P.’s are the same.

∴ t_n1 = t_n2

⇒ 6n – 5 = 105 – 5n

⇒ 6n + 5n = 105 + 5

⇒ 11n = 110

⇒ n = 110/11 = 10

∴ For the nth terms of the two given A.P.’s to be same, the value of n = 10.

Question 10.

How many two digit numbers are divisible by 13?

Answer:

The two digits form the A.P.: 10, 11, 12 … 99.

The two digit numbers that are divisible by 13 form the A.P.: 13, 26, 39 … 91.

Here, first term, a = 13

We know that common difference, d = a₁ – a.

∴ d = 26 – 13 = 13

And last term, l = 91

We know that l = a + (n – 1) d.

⇒ 91 = 13 + (n – 1) (13)

⇒ 91 – 13 = (n – 1) (13)

⇒ 78 = (n – 1) (13)

⇒ 78/13 = (n – 1)

⇒ 6 = n – 1

⇒ n = 6 + 1 = 7

∴ There are 7 two digit numbers that are divisible by 13.

Question 11.

A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year and in the 15th year.

Answer:

Here, t₇ = 1000 and t₁₀ = 1450

We know that nth term of A.P. , t_n = a + (n – 1) d.

First, t₇ = a + (7 – 1) d = 1000

⇒ a + 6d = 1000 … (1)

Then, t₁₀ = a + (10 – 1) d = 1450

⇒ a + 9d = 1450 … (2)

From (1) and (2),

a + 6d = 1000

a + 9d = 1450

(-) (-) (-)

-3d = -450

⇒ d = 450/3 = 150

Substituting d = 150 in (1),

⇒ a + 6(150) = 1000

⇒ a = 1000 – 900 = 100

Now, t₁ = 100 + (1 – 1) (150)

= 100 + 0

= 100

And t₁₅ = 100 + (15 – 1) (150)

= 100 + 14 (150)

= 100 + 2100

= 2200

∴ Number of TVs produced in the first year are 100 and in the 15^th year are 2200.

Question 12.

A man has saved ₹640 during the first month, ₹720 in the second month and ₹800 in the third month. If he continues his savings in this sequence, what will be his savings in the 25th month?

Answer:

The savings form the A.P.: 640, 720, 800 …

Here, first term, a = 640

We know that common difference, d = a₁ – a.

∴ d = 720 – 640 = 80

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ t₂₅ = 640 + (25 – 1) (80)

= 640 + (24) (80)

= 640 + 1920

= 2560

∴ The man’s savings in the 25^th month are Rs. 2560.

Question 13.

The sum of three consecutive terms in an A.P. is 6 and their product is –120. Find the three numbers.

Answer:

We know that the three consecutive terms of an A.P. may be taken as a – d, a, a + d.

Given, a – d + a + a + d = 6

⇒ 3a = 6

⇒ a = 6/3 = 2

Also given (a – d) (a) (a + d) = -120

We know that (a – b) (a + b) = a² – b²

⇒ (a² – d²) (2) = -120

⇒ (2² – d²) = -120/2

⇒ 4 – d² = -60

⇒ d² = 60 + 4 = 64

⇒ d = 8

So, the numbers are

⇒ a – d = 2 – 8 = -6

⇒ a = 2

⇒ a + d = 2 + 8 = 10

∴ The three consecutive numbers are -6, 2, 10.

Question 14.

Find the three consecutive terms in an A. P. whose sum is 18 and the sum of their squares is 140.

Answer:

We know that the three consecutive terms of an A.P. may be taken as a – d, a, a + d.

Given, a – d + a + a + d = 18

⇒ 3a = 18

⇒ a = 18/3 = 6

Also given (a – d)² + a² + (a + d)² = 140

We know that (a – b)² = a² – 2ab + b²

And (a + b)² = a² + 2ab + b²

⇒ a² – 2ad + d² + a² + a² + 2ad + d² = 140

⇒ 3a² + 2d² = 140

⇒ 3(6)² + 2d² = 140

⇒ 108 + 2d² = 140

⇒ 2d² = 140 – 108 = 32

⇒ d² = 32/2 = 16

⇒ d = 4

So, the numbers are

⇒ a – d = 6 – 4 = 2

⇒ a = 6

⇒ a + d = 6 + 4 = 10

∴ The three consecutive numbers are 2, 6, 10.

Question 15.

If m times the mth term of an A.P. is equal to n times its nth term, then show that the (m + n)th term of the A.P. is zero.

Answer:

We know that nth term of A.P. , t_n = a + (n – 1) d.

⇒ First, t_n = a + (n – 1) d

⇒ Then, t_m = a + (m – 1) d

Given, mt_m = nt_n

⇒ m [a + (m – 1) d] = n [a + (n – 1) d]

⇒ ma + m²d – md = na + n²d – nd

⇒ ma – na + m²d – n²d – md + nd = 0

⇒ a (m – n) + d (m² – n²) – d (m – n) = 0

We know that a² – b² = (a – b) (a + b)

⇒ (m – n) [a + d (m + n) – d] = 0

⇒ [a + d (m + n) – d] = 0

⇒ a + (m + n – 1) d = 0

∴ (m + n)th term, t_{m + n} = 0

Hence proved.

Question 16.

A person has deposited ₹25,000 in an investment which yields 14% simple interest annually. Do these amounts (principal + interest) form an A.P.? If so, determine the amount of investment after 20 years.

Answer:

Yes, the amounts form an A.P.

Given principal, p = Rs. 25,000

Simple Interest, r = 14%

Time, t = 20 years

We know that Total amount = p (1 + )

⇒ Total Amount = 25000 (1 + )

= 25000 (1 + )

= 25000 ()

= 95000

∴ Amount of investment after 20 years = Rs. 95, 000

Question 17.

If a, b, c are in A.P. then prove that (a – c)² = 4 (b² – ac).

Answer:

Given, a, b and c are in A.P.

We know that when t₁, t₂, t₃ … are in A.P., t₃ – t₂ = t₂ – t₁

⇒ c – b = b – a

⇒ 2b = a + c

Squaring on both sides,

⇒ (2b)² = (a + c)²

We know that (a + b)² = a² + 2ab + b².

⇒ 4b² = a² + 2ac + c²

Subtracting 4ac on both sides,

⇒ 4b² – 4ac = a² + 2ac + c²– 4ac

⇒ 4 (b² – ac) = a² – 2ac + c²

∴ 4 (b² – ac) = (a – c)²

Hence proved.

Question 18.

If a, b, c are in A.P. then prove that  are also in A.P.

Answer:

Given, a, b, c are in A.P.

Here, first term = a

Common difference, d₁ = b – a … (1)

and d₂ = c – b … (2)

Consider,  and,

Common difference, d₃ =  - 

= 

=  … (3)

⇒ d₄ =  - 

= 

=  … (4)

From (1) and (2),

⇒ d₁ = d₂

⇒ b – a = c – b

Dividing both sides by abc,

⇒  = 

⇒ d₃ = d₄ [From (3) and (4)]

Hence,,  and are in A.P.

Question 19.

If a², b², c²are in A.P. then show that are also in A.P.

Answer:

Given, a², b² and c² are in A.P.

We know that when t₁, t₂, t₃ … are in A.P., t₃ – t₂ = t₂ – t₁

⇒ b² – a² = c² – b²

We know that a² – b² = (a – b) (a + b)

⇒ (b – a) (b + a) = (c – b) (c + b)

⇒  = 

Dividing by (c + a) on both sides,

⇒  = 

⇒  = 

⇒ 

Hence, ,  and are in A.P.

Question 20.

If a^x = b^y = c^z, x ≠ 0, y ≠ 0, z ≠ 0 and b² = ac, then show that are in A.P.

Answer:

Let a^x = b^y = c^z = k.

We know that if a^m = k, then a =.

⇒ a = , b = , c = 

Given, b² = ac

⇒  =  × 

We know that (a^m)ⁿ = a^mn and a^m × aⁿ = a^{m + n}.

⇒  = 

Bases are same, so we equate the powers.

⇒  =  + 

⇒  +  =  + 

⇒  -  =  - 

We know that when t₁, t₂, t₃ … are in A.P., t₃ – t₂ = t₂ – t₁

∴,  and  are in A.P.

---

Exercise 2.3

Question 1.

Find out which of the following sequences are geometric sequences. For those geometric sequences, find the common ratio.

0.12, 0.24, 0.48,….

Answer:

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given: a₁ = 0.12, a₂ = 0.24, a₃ = 0.48

Now,

And,

Therefore,

Now, ∵ the ration is same,

⇒ 0.12, 0.24, 0.48, … is a G.P with common ratio = 2

Question 2.

Find out which of the following sequences are geometric sequences. For those geometric sequences, find the common ratio.

0.004, 0.02, 0.1,….

Answer:

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given: a₁ = 0.004, a₂ = 0.02, a₃ = 0.1

Now,

And,

Therefore,

Now, ∵ the ration is same,

⇒ 0.004, 0.02, 0.1, … is a G.P with common ratio = 5

Question 3.

Find out which of the following sequences are geometric sequences. For those geometric sequences, find the common ratio.

….. .

Answer:

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given:

Now,

And,

And,

Therefore,

Now, ∵ the ration is same,

Question 4.

Find out which of the following sequences are geometric sequences. For those geometric sequences, find the common ratio.

Answer:

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given: a₁ = 12, a₂ = 1, 

Now,

And

Therefore,

Now, ∵ the ration is same,

Question 5.

Find out which of the following sequences are geometric sequences. For those geometric sequences, find the common ratio.

Answer:

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given:

Now,

And

Thherefore,

Now, ∵ the ration is same,

Question 6.

Find out which of the following sequences are geometric sequences. For those geometric sequences, find the common ratio.

Answer:

(Note: Sequences are in G.P. if they have common ratios

i.e., if

)

Given: a₁ = 4, a₂ = -2, a₃ = -1,

Now,

And,

And,

Therefore,

Now, ∵ the ration is not same,

Question 7.

Find the 10th term and common ratio of the geometric sequence  .

Answer:

Given:

 ; (where, n = 1,2,……n-1)

Now, taking n = 3.

⇒ r = -2

Also, a_n = a₁r^n-1 (n = no. of term and a₁ = first term of G.P)

∴ 10^th term (a₁₀ = a₁r⁹)

⇒ a₁₀ = -2⁷

⇒ common ratio = -2 and 10^th term = -2⁷

Question 8.

If the 4th and 7th terms of a G.P. are 54 and 1458 respectively, find the G.P.

Answer:

a₄ = 54 and a₇ = 1458

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒ a₄ = a₁r³

⇒ 54 = a₁r³ ………(1)

Also, ⇒ a₇ = a₁r⁶

⇒ 1458 = a₁r⁶ ………(2)

Dividing equation (1) & (2), we get-

⇒ r³ = 27

⇒ r³ = 3³

⇒ r = 3

Now, putting value of r in equation (1), we get-

⇒54 = a₁3³

⇒ a₁ = 2

Now, G.P will be-

⇒ a₁ , a₁r ,a₁r², a₁r³,………

⇒ 2, 2 × 3, 2 × 3², 2 × 3…

⇒ 2, 6, 18, 54,… is the G.P.

Question 9.

In a geometric sequence, the first term is  and the sixth term is , find the G.P.

Answer:

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒ a₆ = a₁r⁵

………(1)

Now, G.P will be-

⇒ a₁, a₁r, a₁r², a₁r³,………

Question 10.

Which term of the geometric sequence,

(i) 5, 2, … , is  (ii) 1, 2, 4, 8,…, is 1024 ?

Answer:

(i) a₁ = 5, a₂ = 2

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒ n-1 = 7

⇒ n = 8

(ii) a₁ = 1, a₂ = 2

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒1024 = 1 × 2^n-1

⇒ 2¹⁰ = 2^n-1

⇒ n-1 = 10

⇒ n = 11

∴ 1024 is the 11^th term of G.P.

Question 11.

If the geometric sequences 162, 54, 18,…. and ,…have their nth term equal, find the value of n.

Answer:

Given:For 1^st G.P-

a₁ = 162 and a₂ = 54,

And for 2^nd G.P-

.

R = 3

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

And ∵ nth term of both G.P are equal,

⇒ a_n = A_n.

⇒ a₁r^n-1 = A₁R^n-1

⇒ n-1 = 4

⇒ n = 5.

Question 12.

The fifth term of a G.P. is 1875. If the first term is 3, find the common ratio.

Answer:

a₁ = 3 and a₅ = 1875

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

⇒ a₅ = a₁r⁴

⇒1875 = 3r⁴ ………(1)

⇒ r⁵ = 625

⇒ r⁵ = 5⁵

⇒ r = 5 is the common ratio.

Question 13.

The sum of three terms of a geometric sequence is  and their product is 1. Find the common ratio and the terms.

Answer:

⇒ second term = a and third term = ar.

(where, r is common ratio)

……..(1)

Also, their product is 1.

⇒ a³ = 1

⇒ a = 1

Substituting a = 1 in equation (1)

⇒ 10 + 10r + 10r² = 39r

⇒ 10r²-29r + 10 = 0

⇒ 10r²-25r-4r + 10 = 0

⇒ 5r(2r-5)-2(2r-5)

⇒ (5r-2) (2r-5) = 0

Now, G.P will be-

Question 14.

If the product of three consecutive terms in G.P. is 216 and sum of their products in pairs is 156, find them.

Answer:

Let the first term of G.P be .

⇒ second term = a and third term = ar.

(where, r is the common ratio)

∵ sum of three terms is 156

⇒ a³ = 216

⇒ a = 6. ……..(1)

Also,sum of their product in pairs is 216.

 ………(2)

Substituting (1) in (2), we get-

⇒ 36(1 + r + r²) = 156r

⇒ 36 + 36r + 36r² = 156r

⇒ 36 -120r + 36r² = 0

⇒ 12(3r²-10r + 3) = 0

⇒ 3r²-10r + 3 = 0

⇒ 3r²-9r-1r + 3 = 0

⇒ 3r(r-3) -1(r-3) = 0

⇒ (3r-1)(r-3) = 0

Now, G.P will be-

⇒18, 6, 2 or 2, 6, 18 are the three consecutive terms.

Question 15.

Find the first three consecutive terms in G.P. whose sum is 7 and the sum of their reciprocals is 

Answer:

⇒ second term = a and third term = ar.

(where, r is the common ratio)

∵ sum of three terms is 7

……..(1)

 ………(2)

Now, ∵ R.H.S of equation (1) & (2) is equal-

⇒ L.H.S of equation (1) = L.H.S of equation (2)

⇒ a² = 4

⇒ a = √4

⇒ a = 2

Now substituting a = 1 in (2), we get-

⇒ 2 + 2r + 2r² = 7r

⇒ 2r²-5r + 2 = 0

⇒ 2r²-4r-r + 2 = 0

⇒ (2r-1)(r-2) = 0

Now, G.P will be-

⇒ 4, 2, 1 or 1, 2, 4 are the three consecutive terms.

Question 16.

The sum of the first three terms of a G.P. is 13 and sum of their squares is 91. Determine the G.P.

Answer:

Let the first term of G.P be a

⇒ second term = ar and third term = ar².

(where, r is the common ratio)

∵ sum of three terms is 13

⇒ a(1 + r + r²) = 13r……..(1)

Also, sum of their squares is 91.

⇒ a² (1 + r² + r⁴) = 91r² ………(2)

Now, Squaring (1) dividing by (2)

⇒ 7( 1 + r² + r) = 13( 1 + r² - r)

⇒ (7 + 7r² + 7r) = ( 13 + 13r² - 13 r)

⇒ 6r² - 20r + 6 = 0

⇒ 2(3r² - 10r + 3) = 0

⇒ 3r² - 10r + 3 = 0

⇒ 3r² - 9r-r + 3 = 0

⇒ (3r – 1)(r – 3) = 0

Substituting r in equation (1), we get-

⇒a(1 + 3 + 9) = 13 × 3

And,

⇒13a = 13

And

⇒ a = 1 and a = 9.

Now, G.P is-

a, ar, ar²,….

⇒ If r = 3 and a = 1 then,

⇒1, 1×3, 1×3², ……

= 1, 3 , 9 are the first three terms.

And, If  and a = 9 then,

= 9, 3, 1 are the first three terms.

⇒ 1,3,9,… … or 9,3,1,… … is the G.P.

Question 17.

If ₹1000 is deposited in a bank which pays annual interest at the rate of 5% compounded annually, find the maturity amount at the end of 12 years.

Answer:

∵ there is annual compounding interest

⇒ consecutive amounts are forming G.P.

Now,a₁₂ = ?

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

Question 18.

A company purchases an office copier machine for ₹50,000. It is estimated that the copier depreciates in its value at a rate of 15% per year. What will be the value of the copier after 15 years?

Answer:

∵ there is annual depreciation at a constant rate per year

⇒ consecutive value are forming G.P.

Now,a₁₅ = ?

∵ a_n = a₁r^n-1 (n = no. of term, a₁ = first term of G.P, r = common ratio)

Question 19.

If a, b, c, d are in a geometric sequence, then show that (a - b + c) (b + c + d) = ab + bc + cd.

Answer:

Proof:∵ a, b, c, d are in G.P

⇒ a = a, b = ar, c = ar²,d = ar³.

⇒ L.H.S = (a - ar + ar²)(ar + ar² + ar³)

= a²r (1 + r² + r⁴ )

And, R.H.S = a²r + a²r³ + a²r⁵

= a²r (1 + r² + r⁴ )

⇒ L.H.S = R.H.S

Hence, proved that-

(a - b + c) (b + c + d) = ab + bc + cd.

Question 20.

If a, b, c, d are in a G.P., then prove that a + b, b + c, c + d, are also in G.P.

Answer:

Proof: ∵ a, b, c, d are in G.P

⇒ a = a, b = ar ,c = ar²,d = ar³.

To prove: a + b, b + c, c + d, are also in G.P, if-

⇒ (a + b) (c + d) = (b + c)²

Now, we need to prove : (a + b) (c + d) = (b + c)²

L.H.S. = (a + ar)(ar² + ar³)

= a(1 + r) ar² (1 + r)

= a²r² (1 + r)²

R.H.S. = (ar + ar²)²

= (ar(1 + r))²

= a²r² (1 + r)²

⇒ L.H.S = R.H.S

Hence, proved that-

(a + b) (c + d) = (b + c)²

⇒ a + b, b + c, c + d, are also in G.P

---

Exercise 2.4

Question 1.

Find the sum of the first (i) 75 positive integers (ii) 125 natural numbers.

Answer:

(i) In the A.P.

First term = 1

No. of terms = 75

Common difference = 1

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 2850

(ii) In the A.P.

First term = 1

No. of terms = 125

Common difference = 1

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 7875

Question 2.

Find the sum of the first 30 terms of an A.P. whose nth term is 3 + 2n.

Answer:

In the A.P.

First term = 3 + 2×1 = 5

No. of terms = 30

Last term = 3 + 2×30 = 63

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 1020

Question 3.

Find the sum of each arithmetic series

(i) 38 + 35 + 32 + … + 2. (ii)  terms.

Answer:

(i) In the A.P.

First term = 38

Last term = 2

Common difference = 35–38 = –3

Nth term = a + (n–1) d

⇒ 2 = 38 + (n–1)(–3)

⇒ –36 = (n–1)(–3)

⇒ 12 = (n–1)

⇒ n = 13

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 260

(ii) In the A.P.

First term = 6

No. of terms = 25

Common difference = 21/4 – 6 = –3/4

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = –75

Question 4.

Find the S_nfor the following arithmetic series described.

(i) a = 5, n = 30, l = 121 (ii) a = 50, n = 25, d = – 4

Answer:

(i) Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 1890

(ii) Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 50

Question 5.

Find the sum of the first 40 terms of the series 1² – 2² + 3² – 4² + … .

Answer:

Let the sum of n terms be,

S_n= 1² –2² +3² –4²+5² –6²+7² –8² ….

= (1² –2²) + (3² –4²) + (5² –6²) + (7²–8²) …..

= (1–4) + (9–16) + (25–36) + (49–64)……..

= –3 –7 –11 –15………………. [No. of terms  ]

This represents an A.P.

(NOTE: Here the number of terms has been halved. If n= no. of terms in original series and m= no. of terms in new A.P. then  )

Now, here a= –3, d= (–7 – (–3)) = –4

Sum of m terms = 

=

=

=

=

=

As n= 40,

Required sum = 

= –820

Question 6.

In an arithmetic series, the sum of first 11 terms is 44 and that of the next 11 terms is 55. Find the arithmetic series.

Answer:

Sum of terms = 

⇒ 44 = 

⇒ 8 = 2a + 10d

⇒ 2a + 10d = 8 ……………………….(1)

Now sum of next 11 terms is 55

Sum of first 22 terms = Sum of first 11 terms + Sum of next 11 terms

⇒ Sum of first 22 terms = 44 + 55 = 99

Sum of terms = 

⇒ 99 = 

⇒ 9 = 2a + 21d

⇒ 2a + 21d = 9 ……….(1)

Subtracting (1) from (2), we get

⇒ 11d = 1

⇒ d = 1/11

Putting value of d in (1), we get

⇒ 2a + 10(1/11) = 8

⇒ 2a = 8–(10/11)

⇒ 2a = 78/11

⇒ a = 39/11

Therefore, the series is 

Question 7.

In the arithmetic sequence 60, 56, 52, 48,…, starting from the first term, how many terms are needed so that their sum is 368?

Answer:

In the A.P.

First term = 60

Common difference = 56 – 60 = –4

Sum of terms = 

⇒ 368 = 

⇒ 368 = 

⇒ 736 = n (124 – 4n)

⇒ 4n² – 124n + 736 = 0

⇒ 4n² – 92n–32n + 736 = 0

⇒ 4n(n–23) –32(n–23) = 0

⇒ (4n–32)(n–23) = 0

⇒ 4n = 32 or n = 23

⇒ n = 8 or n = 23

Therefore, no. of terms can be 8 or 23

Question 8.

Find the sum of all 3 digit natural numbers, which are divisible by 9.

Answer:

Series of three digit numbers divisible by 9 is:

108, 117,………………………………999

In the A.P.

First term = 108

Last term = 999

Common difference = 9

Nth term = a + (n–1) d

⇒ 999 = 108 + (n–1)9

⇒ 891 = (n–1)9

⇒ 99 = (n–1)

⇒ n = 100

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 55350

Question 9.

Find the sum of first 20 terms of the arithmetic series in which 3rd term is 7 and 7th term is 2 more than three times its 3rd term.

Answer:

First term = a

Common difference = d

3^rd term = a + (3–1)d = a + 2d

7^th term = a + (7–1)d = a + 6d

Now, 3^rd term = 7

⇒ a + 2d = 7 ………………………(1)

And, 7^th term = 3 × (3^rd term) + 2

⇒ a + 6d = 3(a + 2d) + 2

⇒ a + 6d = 3a + 6d + 2

⇒ 2a = –2

⇒ a = –1

Putting value of a in (1)

(⇒ –1) + 2d = 7

⇒ 2d = 8

⇒ d = 4

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 740

Question 10.

Find the sum of all natural numbers between 300 and 500 which are divisible by 11.

Answer:

Series of all natural numbers divisible by 11 between 300and 500 is:

308, 319,………………………………495

In the A.P.

First term = 308

Last term = 495

Common difference = 11

Nth term = a + (n–1) d

⇒ 495 = 308 + (n–1)11

⇒ 187 = (n–1)11

⇒ 17 = (n–1)

⇒ n = 18

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 7227

Question 11.

Solve: 1 + 6 + 11 + 16 + ….. + x = 148.

Answer:

In the A.P.

First term = 1

Common difference = 6 – 1 = 5

Sum of terms = 

⇒ 148 = 

⇒ 148 = 

⇒ 296 = n (5n–3)

⇒ 5n² – 3n –296 = 0

⇒ 5n² –40n + 37n –296 = 0

⇒ 5n(n–8) + 37(n–8) = 0

⇒ (n–8)(5n + 37) = 0

⇒ n = 8 or 5n = –37

As negative value of n is not possible, n = 8

⇒ x = 8^th term of the series

⇒ x = 1 + (8–1)5

⇒ x = 1 + 35 = 36

Question 12.

Find the sum of all numbers between 100 and 200 which are not divisible by 5.

Answer:

We first find the sum of all numbers divisible by 5

Series of all natural numbers divisible by 5 between 100and 200 is:

105, 110,………………………………195

In the A.P.

First term = 105

Last term = 195

Common difference = 5

Nth term = a + (n–1) d

⇒ 195 = 105 + (n–1)5

⇒ 90 = (n–1)5

⇒ 18 = (n–1)

⇒ n = 19

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 2850

Sum of 101,102….199 = Sum of 199 natural numbers – Sum of 100 natural numbers

⇒ Sum(101, 102,…. 199) = 

⇒ Sum(101, 102,…. 199) = 19900–5050 = 14850

Sum of all numbers between 100 and 200 not divisible by 5 = 14850–2850 = 12000

Question 13.

A construction company will be penalised each day for delay in construction of a bridge. The penalty will be ₹4000 for the first day and will increase by ₹1000 for each following day. Based on its budget, the company can afford to pay a maximum of ₹1,65,000 towards penalty. Find the maximum number of days by which the completion of work can be delayed

Answer:

The amount to be paid first paid = ₹4000

It increases each day by = ₹1000

Total amount that can be paid = ₹165000

a = 4000, d = 1000, Sum = 165000

Sum of terms = 

⇒ 165000 = 

⇒ 165 = 

⇒ 330 = n (n + 7)

⇒ n² + 7n –330 = 0

⇒ n² –15n + 22n –330 = 0

⇒ n(n–15) + 22(n–15) = 0

⇒ (n–15)(n + 22) = 0

⇒ n = 15 or n = –22

As negative value of n is not possible, n = 15

Therefore, number of days that construction can be delayed = 15days.

Question 14.

A sum of ₹1000 is deposited every year at 8% simple interest. Calculate the interest at the end of each year. Do these interest amounts form an A.P.? If so, find the total interest at the end of 30 years.

Answer:

Simple interest = P* R * T /100

P = ₹1000

R = 8

T = 1

Simple interest = 1000*8*1/100 = 80

Interest at the end of first year = ₹80

For second year,

P = 1000 + 1000 = 2000

R = 8

T = 1

Simple interest = 2000*8*1/100 = 160

Interest at the end of second year = ₹160

Similarly, for third year, P = 1000 + 1000 + 1000 = 3000

SI = 240.

Yes the simple interests form an A.P.

a = 80

d = 160–80 = 80

n = 30

Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 37200

Question 15.

The sum of first n terms of a certain series is given as 3n²– 2n. Show that the series is an arithmetic series.

Answer:

For n = 1,

Sum = 3(1)²–2(1) = 1

Therefore, first term = 1

For n = 2,

Sum = 3(2)² – 2(2) = 12 – 4 = 8

Second term = 8–1 = 7

For n = 3,

Sum = 3(3)² – 2(3) = 21

Third term = 21– 8 = 13

Series : 1, 7, 13…..

This is an arithmetic progression as the difference between two terms is constant.

Common difference = 7–1 = 13–7 = 6

Question 16.

If a clock strikes once at 1 o’clock, twice at 2 o’clock and so on, how many times will it strike in a day?

Answer:

The clock strikes once at 1, twice at 2 ….so it strikes 12times at 12

In a day this striking from 1 to 12 happens twice.

No. of strikes in one turn from 1 to 12 = 1 + 2 + 3 + …..12

⇒ No. of strikes in one turn from 1 to 12 = 

As the striking happens twice in a day,

Total number of strikes in a day = 78 × 2 = 156

Question 17.

Show that the sum of an arithmetic series whose first term is a, second term b and the last is c is equal to 

Answer:

First term = a

Common difference = (b–a)

Last term = c

⇒ c = a + (n–1)(b–a)

⇒ (n–1) = (c–a) / (b–a)

⇒ n = (b + c – 2a) /( b – a)

Sum of terms = 

⇒ Sum of terms = 

Hence proved.

Question 18.

If there are (2n + 1) terms in an arithmetic series, then prove that the ratio of the sum of odd terms to the sum of even terms is (n + 1) : n.

Answer:

In the A.P, let

First term = a

Common difference = d

Number of terms = (2n + 1)

Series: a,a + d,a + 2d……a + 2nd

For Odd terms

: a, a + 2d,…a + 2nd

First term = a

Common difference = 2d

Number of terms = n + 1

Sum of terms = 

Sum of odd terms = 

⇒ Sum of odd terms = 

For Even terms

: a + d, a + 3d,…a + (2n–1)d

First term = a + d

Common difference = 2d

Number of terms = n

Sum of terms = 

Sum of even terms = 

⇒ Sum of even terms = 

Sum of odd terms : Sum of even terms =  : = 

∴ Sum of odd terms : Sum of even terms = (n + 1) : n

Question 19.

The ratio of the sums of first m and first n terms of an arithmetic series is m²: n²show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1)

Answer:

Sum of m terms = 

Sum of n terms = 

Sum of m terms : Sum of n terms = m² : n²

⇒  :  = m² : n²

⇒ n²m(2a + (m–1)d) = nm²( 2a + ( n–1)d)

⇒ 2an²m + n²m²d – n²md = 2anm² + n²m²d – nm²d

⇒ 2anm(n–m) = nmd(n–m)

⇒ 2a = d

mth term : nth term = a + (m–1)d : a + (n–1)d

⇒ mth term : nth term = a + (m–1)2a : a + (n–1) 2a

⇒ mth term : nth term = (2m–1) : (2n–1)

Question 20.

A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of trapezoid needs to start with a row of 97 bricks. Each row must be decreased by 2 bricks on each end and the construction should stop at 25th row. How many bricks does he need to buy?

Answer:

Longest side = 97

Each row decreases by = 2×2 = 4

Number of rows = 25

a = 97, d = –4, n = 25

Sum of n terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 

⇒ Sum of terms = 1225

---

Exercise 2.5

Question 1.

Find the sum of the first 20 terms of the geometric series  .

Answer:

In the G.P.,

First term = 5/2

Common ratio =  = 

Sum of n terms = 

⇒ Sum of 20 terms = 

⇒ Sum of 20 terms = 

Question 2.

Find the sum of the first 27 terms of the geometric series  .

Answer:

In the G.P.,

First term = 1/9

Common ratio =  = 

Sum of n terms = 

⇒ Sum of 27 terms = 

⇒ Sum of 27 terms = 

Question 3.

Find S_nfor each of the geometric series described below.

(i) a = 3, t₈ = 384, n = 8. (ii) a = 5, r = 3, n = 12.

Ans. 765

Answer:

(i) t₈ = 384

⇒ ar^8–1 = 384

⇒ (3)r⁷ = 384

⇒ r⁷ = 128

⇒ r = 2

Sum of n terms = 

⇒ Sum of 8 terms = 

⇒ Sum of 8 terms = 3(255) = 765

(ii) Sum of n terms = 

⇒ Sum of 12 terms = 

⇒ Sum of 12 terms = 

Question 4.

Find the sum of the following finite series

(i) 1 + 0.1 + 0.01 + 0.001 + … + ( 0.1)9 (ii) 1 + 11 + 111 + … to 20 terms.

Answer:

(i) In the G.P.,

First term = 1

Common ratio =  = 0.1

Sum of n terms = 

⇒ Sum of 10 terms = 

⇒ Sum of 10 terms = 

(ii) Series = 1 + 11 + 111 + …..20 terms

Series = [9×(1 + 11 + 111 + …..)] ( Multiplying and dividing by 9)

⇒ Series = [9 + 99 + 999 + …..]

⇒ Series = [(10–1) + (100–1) + (1000–1) + …..]

⇒ Series = [10 + 100 + 1000 + ….. – (20×1)]

⇒ Series = [10 + 100 + 1000 + …..] – 

We find the sum of 10 + 100 + 1000…..20terms as:

First term = 10

Common ratio =  = 10

Sum of n terms = 

⇒ Sum of 20 terms = 

⇒ Sum of 20 terms = 

⇒ Series = [] – 

⇒ Series =  – 

Question 5.

How many consecutive terms starting from the first term of the series

(i) 3 + 9 + 27 + … would sum to 1092 ? (ii) 2 + 6 + 18 + … would sum to 728 ?

Answer:

(i) First term = 3

Common ratio =  = 3

Sum of n terms = 

⇒ Sum of n terms = 

⇒ 1092 = 

⇒ 728 = 

⇒ 729 = 

⇒ n = 6

(ii) First term = 2

Common ratio =  = 3

Sum of n terms = 

⇒ 728 = 

⇒ 728 = 

⇒ 729 = 

⇒ n = 6

Question 6.

The second term of a geometric series is 3 and the common ratio is . Find the sum of

first 23 consecutive terms in the given geometric series.

Answer:

⇒ 

⇒ 

Sum of n terms = 

⇒ Sum of 23 terms = 

⇒ Sum of 23 terms = 

Question 7.

A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is 9 and sum of the last two terms is 36. Find the series.

Answer:

Let the first term = a

And common ratio = r

Series : a, ar, ar², ar³

⇒ a + ar = 9

⇒ a(1 + r) = 9 ………………(1)

⇒ ar² + ar³ = 36

⇒ ar²(1 + r) = 36

⇒ 9r² = 36 ( From Equation (1))

⇒ r² = 4

⇒ r = 2

Putting the value of r in (1),

⇒ a(1 + 2) = 9

⇒ 3a = 9

⇒ a = 3

∴ Series : 3 + 6 + 12 + 24

Question 8.

Find the sum of first n terms of the series

(i) 7 + 77 + 777 + … . (ii) 0.4 + 0.94 + 0.994 + … .

Answer:

(i) Series = 7 + 77 + 777 + …..n terms

Series = [9×(1 + 11 + 111 + …..)] ( Multiplying and dividing by 9)

⇒ Series = [9 + 99 + 999 + …..]

⇒ Series = [(10–1) + (100–1) + (1000–1) + …..]

⇒ Series = [10 + 100 + 1000 + ….. – (n×1)]

⇒ Series = [10 + 100 + 1000 + …..] – 

We find the sum of 10 + 100 + 1000…..n terms as:

First term = 10

Common ratio =  = 10

Sum of n terms = 

⇒ Sum of n terms = 

⇒ Sum of n terms = 

⇒ Series = [] – 

⇒ Series =  – 

(ii) Series = 0.4 + 0.94 + 0.994 + …..n terms

Series = (1–0.6) + (1–0.06) + (1–0.006) + ……..

⇒ Series = n×1 – (0.6 + 0.06 + 0.006 + ….)

We find the sum of 0.6 + 0.06 + 0.006…..n terms as:

First term = 0.6

Common ratio =  = 

Sum of n terms = 

⇒ Sum of n terms = 

⇒ Sum of n terms = 

⇒ Series = n – 

Question 9.

Suppose that five people are ill during the first week of an epidemic and each sick person

spreads the contagious disease to four other people by the end of the second week and so on. By the end of 15th week, how many people will be affected by the epidemic

Answer:

People infected in first week = 5

More people infected by each person = 4

Number of weeks = 15

a = 5, r = 4, n = 15

Sum of n terms = 

⇒ Sum of 15 terms = 

⇒ Sum of 15 terms = 

So, Number of people infected by 15^th week = 

Question 10.

A gardener wanted to reward a boy for his good deeds by giving some mangoes. He gave the boy two choices. He could either have 1000 mangoes at once or he could get 1 mango on the first day, 2 on the second day, 4 on the third day, 8 mangoes on the fourth day and so on for ten days. Which option should the boy choose to get the maximum number of mangoes?

Answer:

In the second option,

Number of mangoes on first day = 1

Number of times of mangoes on subsequent day = 2

Number of days = 10

a = 1, r = 2, n = 10

Sum of n terms = 

⇒ Sum of 10 terms = 

⇒ Sum of 10 terms =  = 1023

Therefore, the boy will get more mangoes in 2^nd case as there were only 1000 mangoes in the 1^st case.

Question 11.

A geometric series consists of even number of terms. The sum of all terms is 3 times the sum of odd terms. Find the common ratio.

Ans. r = 2

Answer:

In the G.P.,

Let First term = a,

Common ratio = r

Series: a, ar, ar²,…….ar^n–1

Sum of all terms = 

For odd terms,

a, ar²,………ar^n–2

First term = a

Common ratio = r²

Number of terms = n/2

Sum of odd terms = 

⇒ Sum of odd terms = 

Now,

Sum of all terms = 3× Sum of odd terms

⇒  = 3 × 

⇒ (1–r2) = 3(1–r)

⇒ r2–3r + 2 = 0

⇒ r2–2r–r + 2 = 0

⇒ r(r–2)–1(r–2) = 0

⇒ (r–1) (r–2) = 0

r = 1 or r = 2

But r = 1 is not possible, So r = 2.

Question 12.

If S₁,S₂and S₃ are the sum of first n, 2n and 3n terms of a geometric series respectively,

then prove that S₁(S₃ – S₂) = (S₂ – S₁)².

Answer:

Sum of n terms = 

S₁ = 

S₂ = 

S₃ = 

Putting value of S₁, S₂ and S₃ on the left side, we get:

S₁(S₃ – S₂) = 

⇒ S₁(S₃ – S₂) = 

⇒ S₁(S₃ – S₂) = 

⇒ S₁(S₃ – S₂) = 

⇒ S₁(S₃ – S₂) = ………..(1)

Now, we solve the right side by putting S₁, S₂ and S₃ :
(S₂– S₁)² = 

⇒ (S₂– S₁)² =  ………….(2)

From (1) and (2), we have:

Left hand side = Right Hand side

Hence Proved.

---

Exercise 2.6

Question 1.

Find the sum of the following series.

1 + 2 + 3 + …. + 45

Answer:

Given that series S = 1 + 2 + 3 + 4 + 5… + 45, it has n = 45 terms and to find the sum S.

Formula for sum of first n numbers is

S = 

45x23

= 1035

The sum S = 1 + 2 + 3 + … + 45 = 1035

Question 2.

Find the sum of the following series.

16² + 17² + 18 + … + 25²

Answer:

Given the the series S = 16² + 17² + 18 + … + 25²

Let S₁ = 1² + 2² + 3² + … + 25² with n = 25

Let S₂ = 1² + 2² + 3² + .. + 15² with n = 15

Required sum S = S₁-S₂

To find the sum S₁ -

Formula to find the sum of first n squares of natural numbers is

S = 

 S₁ = 

= 

S₁ = 5525

To find the sum S₂ -

Formula to find the sum of first n squares of natural numbers is

S = 

 S₂ = 

S₂ = 

S₂ = 1240

Required sum S = S₁-S₂

S = 5525-1240 = 4285

The sum S = 16² + 17² + 18 + … + 25² = 4285

Question 3.

Find the sum of the following series.

2 + 4 + 6 + … + 100

Answer:

Given the series S = 2 + 4 + 6 + … + 100,

We see that there is a common term 2 in all the numbers in the series. Taking 2 common, we have

S = 2(1 + 2 + 3 + .. + 50)

Let 1 + 2 + 3 + .. + 50 be S₁, with n = 50

 S = 2S₁

To find S₁ -

Formula for sum of first n numbers is

S₁ = 

25x51

= 1275

S = 2S₁

= 2 × 1275 = 2550

S = 2550

The sum S = 2 + 4 + 6 + … + 100 = 2550

Question 4.

Find the sum of the following series.

7 + 14 + 21…. + 490

Answer:

Given the series S = 7 + 14 + 21…. + 490,

We see that there is a common term 7 in all the numbers in the series. Taking 7 common, we have

S = 7(1 + 2 + 3 + .. + 70)

Let 1 + 2 + 3 + .. + 70 be S₁, with n = 70

So S becomes S = 7S₁

To find S₁-

Formula for sum of first n numbers is

S₁ = 

35 × 71

= 2485

S = 7S₁

S = 7 × 2485 = 17395

The sum S = 7 + 14 + 21…. + 490 = 17395

Question 5.

Find the sum of the following series.

5² + 7² + 9² + … + 39²

Answer:

Given the the series S = 5² + 7² + 9² + … + 39²

Let S₁ = 1² + 3²

Let S₂ = 2² + 4² + 6² + .. + 38² with n = 19

Let S₃ = 1² + 2² + 3² + 4² + .. + 39² with n = 39

Required sum S = S₃-S₂-S₁

To find the sum S₁ -

S₁ = 1 + 9 = 10

To find the sum S₂ –

S2 can be rewritten as S₂ = 2²(1² + 2² + 3² + 4² + .. + 19²)

Formula to find the sum of first n squares of natural numbers is

S = 

S₂ = 

= 

= 

S₂ = 9880

To find the sum S₃ -

Formula to find the sum of first n squares of natural numbers is

S = 

 S₃ = 

= 

= 20540

Required sum S = S₃-S₂-S₁

S = 20540-9880-10 = 10650

The sum S = 5² + 7² + 9² + … + 39² = 10650

Question 6.

Find the sum of the following series.

16³ + 17³ + … + 35³

Answer:

Given the the series S = 16³ + 17³ + … + 35³

Let S₁ = 1³ + 2³ + 3³ + … + 35³ with n = 35

Let S₂ = 1³ + 2³ + 3³ + .. + 15³ with n = 15

Required sum S = S₁-S₂

To find the sum S₁ -

Formula to find the sum of first n cubes of natural numbers is

S = 

 S₁ = 

= 

S₁ = 396900

To find the sum S₂ -

Formula to find the sum of first n cubes of natural numbers is

S = 

 S₂ = 

= 

S₂ = 14400

Required sum S = S₁-S₂

S = 396900-14400 = 382500

The sum S = 16³ + 17³ + … + 35³ = 382500

Question 7.

Find the value of k if

1³ + 2³ + 3³ + … + k³ = 6084

Answer:

Given the first k cubes of natural numbers and that their sum is 6084,

By formula we have S = 

6084 = 

Taking square root on both sides, we get  = 78

k(k + 1) = 156

k² + k = 156

Or k² + k-156 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -156

We get,

k =  is invalid because it yields a negative k which doesn’t make sense because number of terms in a series cannot be negative.

 = 12

The sum S = 1³ + 2³ + 3³ + … + k³ = 6084 corresponds to k = 12.

Question 8.

Find the value of k if

1³ + 2³ + 3³ + … + k³ = 2025

Answer:

Given the first k cubes of natural numbers and that their sum is 2025,

By formula we have S = 

 2025 = 

Taking square root on both sides, we get  = 45

k(k + 1) = 90

k² + k = 90

Or k² + k-90 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -90

We get,

k =  is invalid because it yields a negative k which doesn’t make sense because number of terms in a series cannot be negative.

 = 9

The sum S = 1³ + 2³ + 3³ + … + k³ = 2025 corresponds to k = 9.

Question 9.

If 1 + 2 + 3 + … + p = 171, then find 1³ + 2³ + 3³ + ... + p³.

Answer:

Given that the series S = 1 + 2 + 3 + … + p = 171

We have S = 

p(p + 1) = 342

p² + p = 342

Or p² + p-342 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -342

We get,

p =  is invalid because it yields a negative p which doesn’t make sense because number of terms in a series cannot be negative.

 = 18

S = 1³ + 2³ + 3³ + ... + p³ where p = 18

Formula to find the sum of first n cubes of natural numbers is

S = 

 S = 

S = 

S = 29241

The sum S = 1³ + 2³ + 3³ + ... + p³ corresponds to p = 18 and S = 29241.

Question 10.

If 1³ + 2³ + 3³ + … + k³ = 8281, then find 1 + 2 + 3 + … + k.

Answer:

Given that the series S = 1³ + 2³ + 3³ + … + k³ = 8281

Formula to find the sum of first k cubes of natural numbers is

S = 

 8281 = 

Taking square root on both sides,

91 = 

k(k + 1) = 182

k² + k = 182

Or k² + k-182 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -182

We get,

k =  is invalid because it yields a negative k which doesn’t make sense because number of terms in a series cannot be negative.

 = 13

The sum S = 1³ + 2³ + 3³ + ... + k³ corresponds to k = 13.

Given series 1 + 2 + 3…. + k, we have k = 13

We have S = 

= 

= 

S = 91

 The sum S = 1³ + 2³ + 3³ + ... + k³ corresponds to k = 13 and 1 + 2 + 3…. + k = 91.

Question 11.

Find the total area of 12 squares whose sides are 12 cm, 13cm, g, 23cm. respectively.

Answer:

Given that there are 12 squares,

We see that their sides 12cm, 13cm..,23cm are in series.

To find the total area, we know that the area of a square is simply l² where l is the length of the side.

Area of first square = 12cm × 12cm = 12² cm²

Area of the second square = 13cm × 13cm = 13² cm²

And so on.

We observe that this is in a series.

So S = 12² + 13² + 14² + … + 23²

To find S,

Let S₁ be 1² + 2² + …. + 23² with n = 23

Let S₂ be 1² + 2² + …. + 11² with n = 11

S = S₁-S₂

To find S₁

1² + 2² + …. + 23² with n = 23

Formula to find the sum of first n squares of natural numbers is

S = 

 S₁ = 

= 

S₁ = 4324

To find S₂

1² + 2² + …. + 11² with n = 11

Formula to find the sum of first n squares of natural numbers is

S = 

 S₂ = 

= 

S₂ = 506

S = S₁-S₂

S = 4324-506

S = 3818cm²

The required sum S = 12² + 13² + 14² + … + 23² = 3818cm²

Question 12.

Find the total volume of 15 cubes whose edges are 16 cm, 17 cm, 18 cm, …., 30 cm respectively

Answer:

Given that there are 15 cubes,

We see that their sides 16cm, 17cm..,30cm are in series.

To find the total volume, we know that the volume of a cube is simply l³ where l is the length of the side.

Volume of first cube = 16cm × 16cm × 16cm = 16³ cm³

Volume of second cube = 17cm × 17cm × 17cm = 17³ cm³

And so on.

We observe that this is in a series.

So S = 16³ + 17³ + 18³ + … + 30³

To find S,

Let S₁ be 1³ + 2³ + …. + 30³ with n = 30

Let S₂ be 1³ + 2³ + …. + 15³ with n = 15

S = S₁-S₂

To find S₁

1³ + 2³ + …. + 30³ with n = 30

Formula to find the sum of first n cubes of natural numbers is

S = 

 S₁ = 

= 

S₁ = 216225

To find the sum S₂ –

1³ + 2³ + …. + 15³ with n = 15

Formula to find the sum of first n cubes of natural numbers is

S = 

 S₂ = 

= 

S₂ = 14400

S = S₁-S₂

S = 216225-14400

S = 201825 cm³

The required sum S = 16³ + 17³ + 18³ + … + 30³ = 201825 cm³

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Exercise 2.7

Question 1.

Which one of the following is not true?
A. A sequence is a real valued function defined on N.

B. Every function represents a sequence.

C. A sequence may have infinitely many terms.

D. A sequence may have a finite number of terms.

Answer:

“Not true” – tells us that there is one option among A, B C or D which is false, while the rest are true.

Let us examine each option separately.

● A, is true because a sequence a is defined on a set of natural numbers.

● C, is true because an infinite sequence is possible. (ex. Infinite GP)

● D, is true because a finite sequence is possible. (ex. Finite AP, GP or HP)

● B, is false because every function is not a sequence, but every sequence is a function.

Hence, the correct option is B.

Question 2.

The 8th term of the sequence 1, 1, 2, 3, 5, 8, g is
A. 25

B. 24

C. 23

D. 21

Answer:

The above series is also called a Fibonacci sequence, where each term is the sum of its preceding two terms.

a_n = a_n-1 + a_n-2

We are interested in 8^th term so substituting n = 8 in the above equation,

a₈ = a₇ + a₆. Also, a₇ = a₆ + a₅

 a₈ = (a₆ + a₅) + a₆

 a₈ = (8 + 5) + 8

 a₈ = 21

So the correct option is D.

Question 3.

The next term of  in the sequence is
A. 

B. 

C. 

D. 

Answer:

The general term would be

Where n = 1,2,3….

Next term after is the fifth term so n = 5

We have

 = 

Therefore, the correct option is C.

Question 4.

If a, b, c, l, m are in A.P, then the value of a - 4b + 6c - 4l + m is
A. 1

B. 2

C. 3

D. 0

Answer:

If a, b, c, l, m are in A.P, then

b-a = c-b = l-c = m-l.

a-4b + 6c-4l + m becomes

a-b-b-b-b + c + c + c + c + c + c-l-l-l-l + m

Grouping the terms

We have

a + 4(c-b)-2(l-c)-2l + m

a + 2(c-b)-2(l-c) + 2(c-b)-2l + m

Since c-b = l-c,

a + 2(c-b)-2l + m

a + 2c-2b-2l + m

-(b-a) + (c-b)-(l-c) + (m-l)

= 0

So, the correct option is D.

Question 5.

If a, b, c are in A.P. then  is equal to
A. 

B. 

C. 

D. 1

Answer:

If a,b,c are in AP, then b-a = c-b

Or 

Or 

Therefore, the correct option is D.

Question 6.

If the n^th term of a sequence is 100 n + 10, then the sequence is
A. an A.P.

B. a G.P.

C. a constant sequence

D. neither A.P. nor G.P.

Answer:

given nth term = 100n + 10

This can be rewritten as 110 + (n-1)100

This is in the form T_n = a + (n-1)d which forms an AP.

Therefore, the correct option is A.

Question 7.

If a₁, a₂ , a₃,...are in A.P. such that  , then the 13th term of the A.P. is
A. 

B. 0

C. 12_a1

D. 14_a1

Answer:

Given 

Let common ratio be d

a₄ = a₁ + 3d

a₇ = a₁ + 6d

 2a₁ + 6d = 3a₁ + 18d

 a₁ + 12d = 0

Which corresponds to the 13^th term (a₁ + (13-1)d) = 0

Therefore, the correct option is B.

Question 8.

If the sequence a₁, a₂, a₃,… is in A.P. , then the sequence a₅, a₁₀, a₁₅,…is
A. a G.P.

B. an A.P.

C. neither A.P nor G.P.

D. a constant sequence

Answer:

Terms collected from an AP with a common interval between two terms is always in AP, a₅, a₁₀, a_15…. have a common interval of 5, so it is also in AP.

Question 9.

If k + 2, 4k–6, 3k–2 are the three consecutive terms of an A.P, then the value of k is
A. 2

B. 3

C. 4

D. 5

Answer:

From AP, it is clear that

(4k-6)-(k + 2) = (3k-2)-(4k-6)

3k-8 = -k + 4

4k = 12 or k = 3.

So the correct choice is B.

Question 10.

If a, b, c, l, m. n are in A.P., then 3a + 7, 3b + 7, 3c + 7, 3l + 7, 3m + 7, 3n + 7 form
A. a G.P.

B. an A.P.

C. a constant sequence

D. neither A.P. nor G.P

Answer:

Terms of an AP if multiplied with a constant will still be in AP.

The constant term is 3.

Therefore, 3a, 3b, 3c…are in AP.

Adding a constant to an AP does not change the sequence type. i.e. It will still be in AP.

The constant term added is 7.

Therefore, 3a + 7, 3b + 7….also form an AP.

The correct choice is B.

Question 11.

If the third term of a G.P is 2, then the product of first 5 terms is
A. 5²

B. 2⁵

C. 10

D. 15

Answer:

2 = ar^n-1 = ar^3-1 = ar²

Product of the first five terms = axarxar²xar³xar⁴

This can be rewritten as a⁵r¹⁰ or (ar²)⁵

So, Product of the first five terms is 2⁵

Hence the correct option is B.

Question 12.

If a, b, c are in G.P, then is equal to  is equal to
A. 

B. 

C. 

D. 

Answer:

Given that a, b, c are in GP,

Subtracting 1 from both sides, we get

Therefore, the correct option is A.

Question 13.

If x, 2x + 2, 3x + 3 are in G.P, then 5x, 10x + 10, 15x + 15 form
A. an A.P.

B. a G.P.

C. a constant sequence

D. neither A.P. nor a G.P.

Answer:

Any GP if multiplied with a constant term remains in GP.

x, 2x + 2, 3x + 3 are in GP, 5(x), 5(2x + 2), 5(3x + 3) will also form a GP because a constant term 5 is merely multiplied.

So the correct option is B.

Question 14.

The sequence –3, –3, –3,… is
A. an A.P. only

B. a G.P. only

C. neither A.P. nor G.P

D. both A.P. and G.P.

Answer:

The given sequence is -3, -3, -3

Test for AP

If a,b,c…are in AP the b-a = c-b…

-3-(-3) = -3-(-3) = 0

Therefore, the series is in AP.

Test for GP

If a,b,c.. are in GP then …

Therefore, the series is in GP.

The series is both in AP and GP.

So the correct option is D.

Question 15.

If the product of the first four consecutive terms of a G.P is 256 and if the common ratio is 4 and the first term is positive, then its 3rd term is
A. 8

B. 

C. 

D. 16

Answer:

Given the product of first four consecutive terms of GP = 256.

Let the first four terms be a, ar, ar², ar³

Product = 256 = axarxar²xar³ = a⁴r⁶

Also given r = 4,

We have 256 = a⁴ × (4)⁶

a⁴ = 

a =  which is positive.

Third term is ar² =  = 8

So the correct option is A.

Question 16.

In a G.P, t₂ =  and t₃ = . Then the common ratio is
A. 

B. 

C. 1

D. 5

Answer:

Common ratio =  = 

So the correct option is B.

Question 17.

If x 0, then 1 + sec x + sec²x + sec³x + sec⁴x + sec⁵x is equal to
A. (1 sec x) (sec²x + sec³x + sec⁴x)

B. (1 + sec x) (1 + sec²x + sec⁴x)

C. (1 sec x) (sec x + sec³x + sec⁵x)

D. (1 + sec x) (1 + sec³x + sec⁴x)

Answer:

We see that the problem under consideration is a GP with a = 1 and common ratio r = sec x.

Sum of GP S = 

Where n = number of terms, here n = 6.

Therefore

S = 

The numerator is of the form a⁶-b⁶

a⁶-b⁶ = (a + b)(a-b)(a² + b² + ab)(a² + b²-ab)

Therefore (sec x)⁶ -1⁶ = (sec x + 1)(sec x-1)(sec²x + 1 + sec x)(sec²x + 1-sec x)

(sec x)⁶ -1⁶ = (sec x + 1)(sec x-1)(1 + sec²x + sec⁴x)

 S =  = (sec x + 1)(1 + sec²x + sec⁴x)

So the correct answer is option B.

Question 18.

If the n^thterm of an A.P. is t_n = 3-5n, then the sum of the first n terms is
A. 

B. n(1 - 5n)

C. 

D. 

Answer:

Given t_n = 3-5_n, t₁ = 3-5 = -2

Sum of n terms of an AP = S_n = a + t_n)

S_n = -2 + 3-5n) = 1-5n)

So the correct answer is B.

Question 19.

The common ratio of the G.P. a^m-n, a^m, a^{m + n}is
A. a^m

B. a^-m

C. aⁿ

D. a ^-n

Answer:

If x, y, z are in GP, then Common ratio r =  or 

Common ratio in this problem is  or  = aⁿ

Therefore, the correct option is C.

Question 20.

If 1 + 2 + 3 + . . . + n = k then 1³ + 2³ + … + n³is equal to
A. k²

B. k³

C. 

D. (k + 1)³

Answer:

Sum of first n natural numbers is S =  = k

Sum of first n natural squares is S = 

Which is k²

So the correct option is A.