If { a i } i = 1 n , where n is an even integer, is an arithmetic progression with common difference 1 and i = 1 n a i = 192 , i = 1 n / 2 a 2 i = 120 , then n is equal to

Question

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

Smart Achievers · Accepted Answer

(2)

$\sum_{i = 1}^{n} a_{i} = \frac{n}{2} {2 a_{1} + (n + 1)} = 192$

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

$\sum_{i = 1}^{n / 2} a_{2 i} = \frac{n}{4} [2 a_{1} + 2 + (\frac{n}{2} - 1) 2] = 120$

$2 a_{1} + n = \frac{480}{n} \dots (2)$

$From equation (2) and (1)$

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

$n = 480 - 384 = 96$

Solution

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

1 48

2 96

3 92

4 104

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