If a i i = 1 n , where n is an even integer, is an arithmetic progression with common difference 1 , and…

Given, $\sum _{i=1}^n{a}_i=192$

$a_1+a_2 \cdots a_n=192$

$\Rightarrow \frac{n}{2}\left(a_1+a_n\right)=192$

$a_1+a_n=\frac{384}{n} \cdots \left(1\right)$

Also given $\sum _{i=1}^{\frac{n}{2}}{a}_{2i}=120$

$\Rightarrow \underset{\frac{n}{2} \mathrm{terms}}{\underbrace{a_2+a_4+a_6 \cdots a_n }}=120$

$\Rightarrow \frac{n}{2}\times \frac{1}{2}\left[a_2+a_n\right]=120$

$a_2+a_n=\frac{480}{n}$

$a_1+1+a_n=\frac{480}{n} \cdots \left(2\right)$

Now equation $\left(2\right)-$equation $\left(1\right)$

$\frac{480}{n}-\frac{384}{n}=\left(a_1+a_n+1\right)-\left(a,+a_n\right)$

$\frac{1}{n}\left(480-384\right)=1$

$480-384=n$

$n=96$