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

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Sequences and Series

{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_{2 i} = 120

, then n is equal to:

Answer: (B)

Let the first term be

a_{1}

and common difference

d = 1

. From

\sum_{i = 1}^{n} a_{i} = 192

, we have

\frac{n}{2} [2 a_{1} + (n - 1)] = 192

, giving

2 a_{1} + (n - 1) = \frac{384}{n}

... (1).

The sum of even-indexed terms

\sum_{i = 1}^{n /2} a_{2 i}

forms an AP with first term

a_{2} = a_{1} + 1

and common difference 2. Thus

\frac{n}{4} [2 (a_{1} + 1) + (\frac{n}{2} - 1) \cdot 2] = 120

, simplifying to

2 a_{1} + n = \frac{480}{n}

... (2).

Subtracting (1) from (2):

1 = \frac{480}{n} - \frac{384}{n} = \frac{96}{n}

, so

n = 96

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option