Question

How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?

Answer 1

9 + 17 + 25 + ……. + a_n = 636 

a = 9, d = 17 – 9 = 8, S_n= 636, n =?

n(8n + 10) = 636 × 2 

8n² + 10n = 1272 

8n²+ 10n – 1272 = 0 

4n² + 5n – 636 = 0 

4n²+ 5n – 48n – 636 = 0

n(4n + 5) – 12(4n + 5) = 0 

(4n + 5) (n – 12) = 0 

if n – 12 = 0 then, n = 12 

∴ n = 12.