Question

If
\(\{a_i\}_{i=1}^{n}\)
, where n is an even integer, is an arithmetic progression with common difference 1, and
\(\sum_{i=1}^{n} a_i = 192,\ \sum_{i=1}^{n/2} a_{2i} = 120\)
, then n is equal to :

• A. 48
• B. 96
• C. 92
• D. 104

Hint:

Each odd-indexed term is exactly 1 less than the following even-indexed term, so the difference of the two partial sums equals n/2.

Answer 1

The Correct Option is B

Sum of odd-indexed terms \(= 192 - 120 = 72\).
The terms \(a_{2i}\) and \(a_{2i-1}\) differ by the common difference 1, so \(\sum a_{2i} - \sum a_{2i-1} = \frac{n}{2}\times 1\).
\(120 - 72 = \frac{n}{2} \Rightarrow \frac{n}{2} = 48 \Rightarrow n = 96\).
Answer: (2) 96