Question

Let the sequence a₁, a₂, a₃.....a_n form an A.P. then

$a _ { 2 n - 1 } ^ { 2 } + a _ { 2 n } ^ { 2 }$ is equal to-

• A. $\frac { \mathrm { n } } { 2 \mathrm { n } - 1 } \left( \mathrm { a } _ { 1 } ^ { 2 } - \mathrm { a } _ { 2 \mathrm { n } } ^ { 2 } \right)$
• B. $\frac { 2 n } { n - 1 } \left( a _ { 2 n } ^ { 2 } - a _ { 1 } ^ { 2 } \right)$
• C.
• D. None of these

Answer 1

Answer: (A)

$\frac { \mathrm { n } } { 2 \mathrm { n } - 1 } \left( \mathrm { a } _ { 1 } ^ { 2 } - \mathrm { a } _ { 2 \mathrm { n } } ^ { 2 } \right)$

Answer 2

+

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

= – d(a₁ + a₂ + .......a_2n)

= – d a_2n = a₁ + (2n – 1) d

= – n(a₁ + a_2n) × d =

= .