1 , 7 ,13 ,19, ………………. and 100 , 95 , 90 ,………..
Solution:
tn = a + (n - 1) d
nth term of the first sequence
a = 1 d = t₂-t₁
= 7-1
d = 6
tn = 1 + (n-1) 6
tn = 1 + 6 n – 6
tn = 6 n – 5 -----(1)
nth term of the second sequence
a = 100 d = t₂-t₁
= 95 - 100
d = -5
tn = 100 + (n-1) (-5)
tn = 100 - 5 n + 5
tn = 105 - 5 n -----(2)
(1) = (2)
6 n – 5 = 105 – 5 n
6 n + 5 n = 105 + 5
11 n = 110
n =110/11
n = 11
Therefore 11th terms of the given sequence are equal.