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) $\frac{n+1}{n}$
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).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.