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.