QJEEM22Ma5dbe: JEE Main 2022 Mathematics - Class 11 JEEM22Ma5dbe Solution: If , where n is an even integer, is an arithmetic progression…

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

$\Rightarrow$ 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