If the sums of n terms of two arithmetic progressions are in the ratio

Question:

If the sums of $n$ terms of two arithmetic progressions are in the ratio $\frac{3 n+5}{5 n-7}$, then their $n^{\text {th }}$ terms are in the ratio

(a) $\frac{3 n-1}{5 n-1}$

(b) $\frac{3 n+1}{5 n+1}$

(c) $\frac{5 n+1}{3 n+1}$

(d) $\frac{5 n-1}{3 n-1}$

Solution:

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

$\frac{S_{z}}{S_{n}^{\dagger}}=\frac{3 n+5}{5 n+7}$ ....(1)

We need to find the ratio of their nth 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,

$\frac{[2 a+(n-1) d]}{\left[2 a^{\prime}+(n-1) d^{\prime}\right]}=\frac{3 n+5}{5 n+7}$

Now, to find the ratio of the $n^{\text {th }}$ term, we replace $n$ by $2 n-1$. We get,

$\frac{[2 a+(2 n-1-1) d]}{\left[2 a^{\prime}+(2 n-1-1) d^{\prime}\right]}=\frac{3(2 n-1)+5}{5(2 n-1)+7}$

$\frac{2 a+(2 n-2) d}{2 a^{\prime}+(2 n-2) d^{\prime}}=\frac{6 n-3+5}{10 n-5+7}$

$\frac{2 a+2(n-1) d}{2 a^{\prime}+2(n-1) d^{\prime}}=\frac{6 n+2}{10 n+2}$

$\frac{a+(n-1) d}{a^{1}+(n-1) d^{\prime}}=\frac{3 n+1}{5 n+1}$

As we know,

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

Therefore, we get,

$\frac{a_{n}}{a_{a}^{1}}=\frac{3 n+1}{5 n+1}$

Hence the correct option is (b).

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