Question

If the sequence $a_1,a_2,a_3,.....,a_n$. forms an A.P., then prove that $a_1^2-a_2^2+a_3^2-a_4^2+.....+a_{2n-1}^2-a_{2n}^2=\frac{n}{2n-1}(a_1^2-a_{2n}^2)$.

Answer 1

Given $a_1,a_2,a_3$ are in AP

Let d be the common difference let $d>1$

It means that $a_1-a_2=a_2-a_3=a_3-a_4=...........a_{2n-1}-a_{2n}=-d$

Now let $S=a_1^2-a_2^2+a_3^2-a_4^2.....+a_{2n-1}^2-a_{2n}^{2}S=(a_1+a_2)(a_1-a_2)+(a_3+a_4)(a_3-a_2).......+(a_{2n-1}+a_{2n})(a_{2n-1}-a_{2n})S=(a_1+a_2)(-d)+(a_3+a_4)(-d)....+(a_{2n-a}+a_{2n})(-d)S=(-d)[a_1+a_2+a_3+a_4+....a_{2n-1}+a_{2}n]S=(-d)\left[\frac{2n}{2}\right(a_1+a_{2n}\left)\right]$

{Sum of $2n$ terminal $=\frac{2n}{2}$ first term}

$S=(-d)n(a_1+a_{2n})=(-d)\left[n\right(a_1+a_2\left)\right]$

Since we can write

$a_1-a_2=(a_1-a_2)+(a_2-a_3)+....+(a_{2n-a}-a_{2n})=(2n-1)(-d)$

Hence $=d=\frac{a_1-a_{2n}}{(2n-1)}$ Now replacing $-d$ by $\frac{a_1-a_{2n}}{2n-1}$

$=(-d)\left[n\right(a_1+a_{2n}\left)\right]$ we get

$S=\left[\frac{a_1-a_{2n}}{2n-1}\right]\left[n\right(a_1+a_2\left)\right]=\frac{n}{2n-1}(a_1^2-a_{2n-1})$