If S1 is the sum of an arithmetic series of ' n ' odd number of terms and S2, the sum of the terms of the series in odd places, then (S1/S2)=

Question

If S1 is the sum of an arithmetic series of ' n ' odd number of terms and S2, the sum of the terms of the series in odd places, then (S1/S2)= (A) (2 n/n+1) (B) (n/n+1) (C) (n+1/2 n) (D) (n+1/n)

Tardigrade Admin · Accepted Answer

Let the odd number of terms of an arithmetic series be

a, a + d, a + 2 d, a + 3 d, a + 4 d, \dots\dots, a + (n - 1) d

Then,

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

S_{2} = a + (a + 2 d) + (a + 4 d) + \dots

to

\frac{n + 1}{2}

terms

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

= \frac{n + 1}{4} (2 a + (n - 1) d)

∴ \frac{S _{1}}{S _{2}} = \frac{2 n}{n + 1}

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

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

$\frac{n}{n + 1}$

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

$\frac{n + 1}{n}$

Q. If S1​ is the sum of an arithmetic series of ' n ' odd number of terms and S2​, the sum of the terms of the series in odd places, then S2​S1​​=

n+12n​

n+1n​

2nn+1​

nn+1​

Let the odd number of terms of an arithmetic series be a,a+d,a+2d,a+3d,a+4d,……,a+(n−1)d Then, S1​=2n​{2a+(n−1)d} S2​=a+(a+2d)+(a+4d)+… to 2n+1​ terms =2×2n+1​[2a+(2n+1​−1)×2d] =4n+1​(2a+(n−1)d) ∴S2​S1​​=n+12n​

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

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

$\frac{n}{n + 1}$

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

$\frac{n + 1}{n}$