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

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9 + 17 + 25 + ……. + an = 636 

a = 9, d = 17 – 9 = 8, Sn= 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.

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