Question

The sum of 'n' terms of two A.P.'s are in the ratio of $\frac{5n+2}{11n-7}$. Find the ratio of their sixth terms.

• A. $32:59$
• B. $1:1$
• C. $2:1$
• D. $1:2$

Answer 1

Answer: (D)

$1:2$

Given:

$\frac{S_1}{S_2}=\frac{5n+2}{11n-7}=\frac{\frac{n}{2}\left[5\right(n-1)+5+2]}{\frac{n}{2}\left[11\right(n-1)+11-7]}=\frac{\frac{n}{2}[2\times 3.5+5(n-1\left)\right]}{\frac{n}{2}[2\times 2+11(n-1\left)\right]}$

$\therefore a_1=3.5$ and $d_1=5$

$\therefore a_2=2$ and $d_2=11$

$\therefore \frac{(t_1{)}_6}{(t_2{)}_6}=\frac{a_1+5d_1}{a_2+5d_2}=\frac{3.5+5\times 5}{2+5\times 11}=\frac{1}{2}$

$(Ans\to D)$