Question

The sum of n terms of two AP's are in the ratio of $(3n+8):(7n+15)$. Find the ratio of their ${12}^{\text{th }}$ terms.

• (A) $\frac{7}{12}$
• (B) $\frac{16}{7}$
• (C) $\frac{7}{16}$
• (D) $\frac{12}{7}$

Answer 1

The Correct Option is C

Explanation:
Let us consider two AP's having first term $a_1$ and $a_2$ and the common difference are $d_1$ and $d_2$ respectively.According to the question,$\frac{{\mathrm{S}}_1}{{\mathrm{S}}_2}=\frac{3\mathrm{n}+8}{7\mathrm{n}+15}$Sum of the first $n$ terms $={\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}-1)\times \mathrm{d}]$$\frac{\frac{\mathrm{n}}{2}[2{\mathrm{a}}_1+(\mathrm{n}-1){\mathrm{d}}_1)]}{\frac{\mathrm{n}}{2}[2{\mathrm{a}}_2+(\mathrm{n}-1){\mathrm{d}}_2)]}=\frac{3\mathrm{n}+8}{7\mathrm{n}+15}$$\frac{[{\mathrm{a}}_1+\left(\frac{\mathrm{n}-1}{2}\right){\mathrm{d}}_1)]}{[{\mathrm{a}}_2+\left(\frac{\mathrm{n}-1}{2}\right){\mathrm{d}}_2]}=\frac{3\mathrm{n}+8}{7\mathrm{n}+15}$So, the ratio of ${12}^{\text{th }}$ term of both $\mathrm{AP}$ is:$\frac{a_1+11d_1}{a_2+11d_2}=\frac{[a_1+\left(\frac{n-1}{2}\right)d_1)]}{[a_2+\left(\frac{n-1}{2}\right)d_2]}=\frac{3n+8}{7n+15}$So, we can write $\frac{\mathrm{n}-1}{2}=11$Therefore,$\mathrm{n}=23$So,$\frac{{\mathrm{a}}_1+11{\mathrm{d}}_1}{{\mathrm{a}}_2+11{\mathrm{d}}_2}=\frac{3\times 23+8}{7\times 23+15}=\frac{77}{176}=\frac{7}{16}$Hence, the correct option is (C).