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

$$\Rightarrow$$ 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