Question

if the ratios of the sum of the first n terms of two A.P is 7n+1:4n+27 find the ratios of their 9th terms

Answer 1

Let’s denote the sums of 2 AP’s as S1andS2,S1andS2,

Given that, S1S2=7n+14n+27,S1S2=7n+14n+27,

Sum of first ‘n’ terms of AP, whose first term is ‘a’ and common difference is ‘d’, is given by-

S=n2[2a+(n−1)d]S=n2[2a+(n−1)d]

Let,

First term and common difference of two AP’shaving sum S1S1 and S2S2 be ‘a’ , ’d’ and ‘A’ , ‘D’ respectively.

Then, S1S2=n2[2a+(n−1)d]n2[2A+(n−1)D],S1S2=n2[2a+(n−1)d]n2[2A+(n−1)D],

=> S1S2=2a+(n−1)d2A+(n−1)DS1S2=2a+(n−1)d2A+(n−1)D

=> S1S2=a+n−12dA+n−12DS1S2=a+n−12dA+n−12D ————-(1)

Now,nth term of an AP having first term ‘a’ and common difference ‘d’ is given as,

Tn=a+(n−1)dTn=a+(n−1)d

Thus, 9th term of the two AP’s having sum S1S1 and S2S2 will be a+8da+8d (Let it be A1A1) and A+8DA+8D (Let it be A2A2) respectively,

A1A2=a+8dA+8DA1A2=a+8dA+8D ————-(2)

Comparing equations (1) and (2), we get-

n−12=8n−12=8 or n=17n=17

Given equation,a+n−12dA+n−12D=7n+14n+27a+n−12dA+n−12D=7n+14n+27

Putting n=17n=17 in the above equation,we will get-

a+17−12dA+17−12D=7(17)+14(17)+27a+17−12dA+17−12D=7(17)+14(17)+27

a+162dA+162D=7(17)+14(17)+27a+162dA+162D=7(17)+14(17)+27

a+8dA+8D=A1A2=7(17)+14(17)+27a+8dA+8D=A1A2=7(17)+14(17)+27

Thus,A1A2=120/95=24/19.

hope this helps!
cheers!! (: