Question

The sum of n terms of two series in A.P. are in the ratio 5n +4 : 9n + 6. Find the ratio of their 13^th terms.

• A. $\frac{129}{131}$
• B. $\frac{127}{132}$
• C. $\frac{125}{134}$
• D. $\frac{121}{139}$

Answer 1

Answer: (A)

$\frac{129}{131}$

Answer 2

Let $a_{1},a_{2}$ be the first terms of two A.P.s and $d_{1},d_{2}$ are their respective common differences.

⇒ $\frac{\frac{n}{2}\left\lbrack 2a_{1} + (n - 1)d_{1} \right\rbrack}{\frac{n}{2}\left\lbrack 2a_{2} + (n - 1)d_{2} \right\rbrack} = \frac{5n + 4}{9n + 6}$

⇒$\frac{a_{1} + \frac{(n - 1)}{2}d_{1}}{a_{2} + \frac{(n - 1)}{2}d_{2}} = \frac{5n + 4}{9n + 6}$ .................. (1)

Now the ratio of 13^th terms = $\frac{a_{1} + 12d_{1}}{a_{2} + 12d_{2}}$

⇒ put $\frac{(n - 1)}{2} = 12$i.e. n = 25 in equation (1)

⇒$\frac{a_{1} + 12d_{1}}{a_{2} + 12d_{2}} = \frac{5(25) + 4}{9(25) + 6} = \frac{129}{231}$