Question

Let a₁, a₂, a₃, ….. form an A.P. then a₁² – a₂² + a₃² –a₄² + ……+ a²_2n–1– a²_2n =

• A. $\frac{n}{2n - 1}(a_{1}^{2} - a_{2n}^{2})$
• B. $\frac{2n}{n - 1}(a_{2n}^{2}–a_{1}^{2})$
• C. $\frac{n}{n + 1}(a_{1}^{2} + a_{2n}^{2})$
• D. None of these

Answer 1

Answer: (A)

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

Answer 2

Let 'd' be common difference of A.P.

($a_{1}^{2} - a_{2}^{2}$) + ($a_{3}^{2} - a_{4}^{2}$)+ …….+ ($a_{2n - 1}^{2} - a_{2n}^{2}$)

= (a₁ –a₂) (a₁ + a₂) + (a₃ – a₄) (a₃ + a₄) +……+ (a_2n–1 – a_2n)

= –d(a₁ + a₂)+(–d) (a₃ + a₄) +..+ (–d) (a_2n–1 +a_2n)

= – d[a₁ + a₂ + a₃ + a₄ +…..+a_2n]

= – d$\left\lbrack \frac{2n}{2}(a_{1} + a_{2n}) \right\rbrack$ $\left\lbrack \begin{matrix} a_{2n} = a_{1} + (2n - 1)d \\ d = \frac{a_{2n} - a_{1}}{2n - 1} \end{matrix} \right.\$

= $\left( \frac{a_{1} - a_{2n}}{2n - 1} \right)$× n × (a₁ + a_2n)

= $\frac{n}{2n - 1}$ ($a_{1}^{2} - a_{2n}^{2}$)