If S1 is the sum of an arithmetic progression of 'n' odd number of terms and

Question:

If $\mathrm{S}_{1}$ is the sum of an arithmetic progression of ' $n$ ' odd number of terms and $\mathrm{S}_{2}$ the sum of the terms of the series in odd places, then $\frac{S_{1}}{S_{2}}=$

(a) $\frac{2 n}{n+1}$

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

(c) $\frac{n+1}{2 n}$

(d) $\frac{n+1}{n}$

Solution:

In the given problem, we are given as the sum of an A.P of ‘n’ odd number of terms and the sum of the terms of the series in odd places.

We need to find $\frac{S_{1}}{S_{2}}$

Now, let a1, a2…. an be the n terms of A.P

Where n is odd

Let d be the common difference of the A.P

Then,

$S_{1}=\frac{n}{2}\left[2 a_{1}+(n-1) d\right]$ ......(1)

And be the sum of the terms of the places in odd places,

Where, number of terms $=\frac{n+1}{2}$

Common difference = 2d

So,

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

$S_{2}=\frac{n+1}{4}\left[2 a_{1}+\left(\frac{n-1}{2}\right) 2 d\right]$

$S_{2}=\frac{n+1}{4}\left[2 a_{1}+(n-1) d\right]$ ......(2)

Now,

$\frac{S_{1}}{S_{2}}=\frac{\frac{n}{2}\left[2 a_{1}+(n-1) d\right]}{\frac{n+1}{4}\left[2 a_{1}+(n-1) d\right]}$

$\frac{S_{1}}{S_{2}}=\frac{4 n}{2(n+1)}$

$\frac{S_{1}}{S_{2}}=\frac{2 n}{n+1}$

Thus, $\frac{S_{1}}{S_{2}}=\frac{2 n}{n+1}$

Therefore, the correct option is (a).