Question

An arithmetic series consists of 2n terms, and the first term equals the value of the common difference. If a new series is formed taking the 1st, 3rd, 5th,... (2n -1)th term of the old series, find the ratio of the sum of the new series to that of the sum of the terms of the old series.

• A. $\frac{n+1}{2(2n+1)}$
• B. $\frac{n}{2n+1}$
• C. $\frac{1}{2}$
• D. $Cannotbedetermined$

Answer 1

The correct option is B $\frac{n}{2n+1}$

Best way to solve this question is to assume a simple AP.
Let the AP be 1, 2, 3, 4, 5, 6, 7, 8 (haing 2n = 8 terms, n = 4)
Sum of the original AP = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36
Sum of a new AP formed by taking odd terms = 1 + 3 + 5 + 7 = 16
Required ratio = $\frac{16}{36}=\frac{4}{9}.$
Checking option (b), by putting n = 4,
$\frac{n}{2n+1}=\frac{4}{2\times 4+1}=\frac{4}{9}.$
Hence, option (b) is correct answer.

Alternatively:
The series consists of 2n terms,
First term = a, common diff = a, no of terms = 2n
Sum of all terms $=\frac{2n}{2}[2a+(2n-1\left)a\right]$ ....(1)
Form the new series taking 1st, 3rd, 5th, (2n-1)th term of old series.
First term = a, common difference = 2a, number of terms = n
Sum of all terms $=\frac{n}{2}[2a+(n-1\left)2a\right]$ ...(2)
Dividing (2) by (1) gets the required ratio.