If a_i_i=1^n, where n is an even integer, is an arithmetic progression with common difference 1, and Σ _i=1^n a_i=192,Σ _i=1^ n 2 a_2i=120, then n is…

Given, $\displaystyle \sum _{i=1}^{n}{a}_{i}=192$i=1∑n​ai​=192

$${a}_{1}+{a}_{2}\cdots {a}_{n}=192$$a1​+a2​⋯an​=192

$$\Rightarrow \frac{n}{2}({a}_{1}+{a}_{n})=192$$⇒2n​(a1​+an​)=192

$${a}_{1}+{a}_{n}=\frac{384}{n}\cdots (1)$$a1​+an​=n384​⋯(1)

Also given $\displaystyle \sum _{i=1}^{\frac{n}{2}}{a}_{2i}=120$i=1∑2n​​a2i​=120

$$\Rightarrow \underset{\frac{n}{2}\mathrm{terms}}{\underset{⏟}{{a}_{2}+{a}_{4}+{a}_{6}\cdots {a}_{n}}}=120$$⇒2n​terms⏟a2​+a4​+a6​⋯an​​​=120

$$\Rightarrow \frac{n}{2}\times \frac{1}{2}[{a}_{2}+{a}_{n}]=120$$⇒2n​×21​[a2​+an​]=120

$${a}_{2}+{a}_{n}=\frac{480}{n}$$a2​+an​=n480​

$${a}_{1}+1+{a}_{n}=\frac{480}{n}\cdots (2)$$a1​+1+an​=n480​⋯(2)

Now equation $(2)-$(2)−equation $(1)$(1)

$$\frac{480}{n}-\frac{384}{n}=({a}_{1}+{a}_{n}+1)-(a,+{a}_{n})$$n480​−n384​=(a1​+an​+1)−(a,+an​)

$$\frac{1}{n}(480-384)=1$$n1​(480−384)=1

$$480-384=n$$480−384=n

$$n=96$$n=96