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6) How many terms are there in the following Arithmetic progressions?

(i)-1,-5/6,-2/3,……………10/3

Solution:

First term = -1

               a = -1

Common difference = t2 – t1

                                    = (-5/6) – (-1)

                                    = (-5/6) + 1

                                    = (-5 + 6)/6

                                    = 1/6

tn = 10/3

tn =  a + (n - 1) d

10/3 = -1 + (n - 1) (1/6)

(10/3) + 1 = (n - 1) (1/6)

13/3 = (n - 1) (1/6)

(13/3)(6/1) = (n - 1)

26 = n -1

26 + 1= n

n = 27

Therefore 27 terms are in the given A.P

(ii)7,13,19,……………205

Solution:

First term = 7

               a = 7

Common difference = t2 – t1

                                    = 13 – 7

                                    = 6

tn = 205

tn =  a + (n - 1) d

205 = 7 + (n - 1) (6)

205 - 7 = (n - 1) (6)

198 = (n - 1) (6)

198/6 = (n - 1)

33 = n -1

33 + 1= n

n = 34

Therefore 34 terms are in the given A.P