Question

Show that a₁, a₂, a₃, … form an A.P. where a_n is defined as a_n = 3 + 4n. Also find the sum of first 15 terms.

Answer 1

a_n = 3 + 4n
a₁ = 3 + 4 x 1 = 3 + 4 = 7
a₂ = 3 + 4 x 2 = 3 + 8 = 11
a₃ = 3 + 4 x 3 = 3 + 12 = 15
a₄ = 3 + 4 x 4 = 3 + 16 = 19
and so on Here, a = 1 and d = 11 – 7 = 4
S₁₅ = `n/(2)[2a + (n - 1)d]`

= `(15)/(2)[2 xx 7 + (15 - 1) xx 4]`

= `(15)/(2)[14 + 14 xx 14]`

= `(15)/(2)[14 + 56]`

= `(15)/(2) xx 70`
= 525.