Question

First term and common difference of an A.P. are 1 and 2 respectively; find S₁₀.

Given: a = 1, d = 2

find S₁₀ = ?

S_n = `n/2 [2a + (n - 1) square]`

S₁₀ = `10/2 [2 xx 1 + (10 - 1)square]`

= `10/2[2 + 9 xx square]`

= `10/2 [2 + square]`

= `10/2 xx square = square`

Answer 1

Given: a = 1, d = 2

find S₁₀ = ?

S_n = `n/2 [2a + (n - 1)bbd]`

S₁₀ = `10/2 [2 xx 1 + (10 - 1)bb2]`

= `10/2[2 + 9 xx bb2]`

= `10/2 [2 + bb18]`

= `10/2 xx bb20` = 100