# The sum of n terms of two A.P.'s are in the ratio

Question:

The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is

(a) $\frac{179}{321}$

(b) $\frac{178}{321}$

(C) $\frac{175}{321}$

(d) $\frac{176}{321}$

Solution:

In the given problem, the ratio of the sum of n terms of two A.P’s is given by the expression,

$\frac{S_{n}}{S_{n}^{1}}=\frac{5 n+9}{9 n+6}$.......(1)

We need to find the ratio of their 18th terms.

Here we use the following formula for the sum of $n$ terms of an A.P.,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

So,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where, a and d are the first term and the common difference of the first A.P.

Similarly,

$S_{n}^{\prime}=\frac{n}{2}\left[2 a^{\prime}+(n-1) d^{\prime}\right]$

Where, a’ and d are the first term and the common difference of the first A.P.

So,

$\frac{S_{n}}{S_{n}^{\prime}}=\frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}\left[2 a^{\prime}+(n-1) d^{\prime}\right]}$

$=\frac{[2 a+(n-1) d]}{\left[2 a^{\prime}+(n-1) d^{\prime}\right]}$ .....(2)

Equating (1) and (2), we get,

$\frac{[2 a+(n-1) d]}{\left[2 a^{\prime}+(n-1) d^{\prime}\right]}=\frac{5 n+9}{9 n+6}$

Now, to find the ratio of the nth term, we replace n by. We get,

$\frac{[2 a+(2 n-1-1) d]}{\left[2 a^{\prime}+(2 n-1-1) d^{\prime}\right]}=\frac{5(2 n-1)+9}{9(2 n-1)+6}$

$\frac{2 a+(2 n-2) d}{2 a^{\prime}+(2 n-2) d^{\prime}}=\frac{10 n-5+9}{18 n-9+6}$

$\frac{2 a+2(n-1) d}{2 a^{\prime}+2(n-1) d^{\prime}}=\frac{10 n+4}{18 n-3}$

$\frac{a+(n-1) d}{a^{\prime}+(n-1) d^{\prime}}=\frac{10 n+4}{18 n-3}$

As we know,

$a_{n}=a+(n-1) d$

Therefore, for the 18th terms, we get,

$\frac{a_{18}}{a_{18}^{\prime}}=\frac{10(18)+4}{18(18)-3}$

$=\frac{184}{321}$

Hence $\frac{a_{18}}{a_{15}^{\prime}}=\frac{184}{321}$

Hence no option is correct.