Question

Let the sequence $a_1,a_2,a_3,\dots ,a_n$ form an AP. Then $a_1^2-a_2^2+a_3^2-\cdots a_{2n-1}^2-a_{2n}^2=$

• A. $\frac{n\left(a_1^2-a_{2n}^2\right)}{2n-1}$
• B. $\frac{2n\left(a_{2n}^2-a_1^2\right)}{n-1}$
• C. $\frac{n\left(a_1^2+a_{2n}^2\right)}{n+1}$
• D. none of these

Answer 1

The correct option is A

$\frac{n\left(a_1^2-a_{2n}^2\right)}{2n-1}$

Explanation for the correct option:

Finding the sequence:

Given $a_1,a_2,a_3,\dots ,a_n$ is the sequence in A.P.

Then the common difference '$d$' in A.P. is defined as,

$\mathbf{\Rightarrow }\mathit{d}\mathbf{=}{\mathit{a}}_{\mathbf{2}}\mathbf{-}{\mathit{a}}_{\mathbf{1}}\mathbf{=}{\mathit{a}}_{\mathbf{4}}\mathbf{-}{\mathit{a}}_{\mathbf{3}}\mathbf{=}\mathbf{\cdots }\mathbf{=}{\mathit{a}}_{\mathbf{2}\mathbf{n}}\mathbf{-}{\mathit{a}}_{\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}}$

To find the value of $a_1^2-a_2^2+a_3^2-\cdots a_{2n-1}^2-a_{2n}^2$.

Finding the value of the above expression using the algebraic identity ${\mathit{a}}^{\mathbf{2}}\mathbf{-}{\mathit{b}}^{\mathbf{2}}\mathbf{=}\left(a-b\right)\left(a+b\right)$ for every two terms.

$\begin{array}{rcl}\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\cdots \left(a_{2n-1}^2-a_{2n}^2\right)& =& \left(a_1-a_2\right)\left(a_1+a_2\right)+\left(a_3-a_4\right)\left(a_3+a_4\right)+\cdots +\left(a_{2n-1}-a_{2n}\right)\left(a_{2n-1}+a_{2n}\right)\\ & =& -\left(a_2-a_1\right)\left(a_1+a_2\right)-\left(a_4-a_3\right)\left(a_3+a_4\right)-\cdots -\left(a_{2n}-a_{2n-1}\right)\left(a_{2n-1}+a_{2n}\right)\end{array}$

Since $d=a_2-a_1=a_4-a_3=\cdots =a_{2n}-a_{2n-1}$ we get,

$\begin{array}{rcl}\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\cdots \left(a_{2n-1}^2-a_{2n}^2\right)& =& -d\left(a_1+a_2\right)-d\left(a_3+a_4\right)-\cdots -d\left(a_{2n-1}+a_{2n}\right)\\ & =& -d\left[a_1+a_2+a_3+a_4+\cdots +a_{2n-1}+a_{2n}\right]\\ & =& -d\left[\frac{2n\left(a_1+a_{2n}\right)}{2}\right]\\ & =& -dn\left(a_1+a_{2n}\right)\end{array}$

No we know that the nth term of an A.P. can be written as,

${\mathit{a}}_{\mathbf{2}\mathbf{n}}\mathbf{=}{\mathit{a}}_{\mathbf{1}}\mathbf{+}\left(2n-1\right)\mathit{d}$

Using the above equation and find '$d$',

$\Rightarrow d=\frac{a_{2n}-a_1}{\left(2n-1\right)}$

Substitute the obtained value of '$d$' in $-dn\left(a_1+a_{2n}\right)$.

$\begin{array}{rcl}\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\cdots \left(a_{2n-1}^2-a_{2n}^2\right)& =& -\left(\frac{a_{2n}-a_1}{2n-1}\right)n\left(a_1+a_{2n}\right)\\ & =& -\frac{-n\left(a_1-a_{2n}\right)\left(a_1+a_{2n}\right)}{2n-1}\\ & =& \frac{n\left(a_1^2-a_{2n}^2\right)}{2n-1}\end{array}$

Therefore, the value of $a_1^2-a_2^2+a_3^2-\cdots a_{2n-1}^2-a_{2n}^2$ is equal to $\frac{n\left(a_1^2-a_{2n}^2\right)}{2n-1}$.

Hence, option (A) is the correct answer.