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

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Accepted Answer

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$\sum\limits_{i = 1}^n {{a_i} = 192}$

$\Rightarrow$ a₁ + a₂ + a₃ + ...... + a_n = 192

$\Rightarrow {n \over 2}[{a_1} + {a_n}] = 192$

$\Rightarrow {a_1} + {a_n} = {{384} \over n}$ ..... (1)

Now, $\sum\limits_{i = 1}^{{n \over 2}} {{a_{2i}} = 120}$

$\Rightarrow$ a₂ + a₄ + a₆ + ...... + a_n = 120

Here total ${n \over 2}$ terms present.

$\therefore$ ${{{n \over 2}} \over 2}[{a_2} + {a_n}] = 120$

$\Rightarrow {n \over 4}[{a_1} + 1 + {a_n}] = 120$

$\Rightarrow {a_1} + {a_n} + 1 = {{480} \over n}$ ..... (2)

Subtracting (1) from (2), we get

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

$\Rightarrow 1 = {{96} \over n}$

$\Rightarrow$ n = 96

"

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

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

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