Question

The 104^th term and 4^th term of an A.P. are 125 and 0. Find the sum of first 35 terms

Answer 1

t₁₀₄ = 125

t₄ = 0

a + (n – 1)d = t_n 

  a     +      103d = 125     ...(1)
  a     +          3d = 0        ...(2)
 (–)   (–)                 (–)   
(1) – (2) ⇒ 100d = 125

d = `125/100 = 5/4`

Substitute d = `5/4` in (2)

`"a" + 3 xx 5/4` = 0

`"a" + 15/4` = 0

 ⇒ a = `-15/4`

∴ S_n = `"n"/2 (2"a" + ("n" - 1)"d")`

S₃₅ = `35/2(2 xx (-15)/4 + 34 xx 5/4)`

= `35/2((-15)/2 + 85/2)`

= `35/2(70/2) = 35/2 xx 35`

= `1225/2`

= 612.5