Question

The ratio of the sum of the first $n$ terms of two different A.P.s is given by $\frac{S_n}{S'_n}=\frac{3n+8}{7n+15}$. The ratio of their $9^{th}$ terms is:

• A. 59/134
• B. 7/16
• C. 35/97
• D. 61/135

Answer 1

Answer: (A)

59/134

Answer 2

The ratio of the sum of the first $n$ terms of two APs is $\frac{S_n}{S'_n} = \frac{2a_1 + (n-1)d_1}{2a'_1 + (n-1)d'_1}$. This can be rewritten as $\frac{a_1 + \frac{n-1}{2}d_1}{a'_1 + \frac{n-1}{2}d'_1}$. The ratio of the $k^{th}$ terms is $\frac{T_k}{T'_k} = \frac{a_1 + (k-1)d_1}{a'_1 + (k-1)d'_1}$. Equating $\frac{n-1}{2} = k-1$ gives $n = 2k-1$. For the $9^{th}$ term ($k=9$), $n = 2(9)-1 = 17$. Substituting $n=17$ into the given ratio of sums $\frac{3n+8}{7n+15}$ gives $\frac{3(17)+8}{7(17)+15} = \frac{51+8}{119+15} = \frac{59}{134}$.