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Question
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Solution
Let there be n terms of this A.P.
For this A.P., a = 9
d = a2 − a1
= 17 − 9
= 8
`S_n = n/2[2a+(n-1)d]`
`636 = n/2[2xx9+(-1)8]`
⇒ 636 = 9n + 4n2 - 4n
⇒ 4n2 + 5n - 636 = 0
⇒ 4n2 + 53n - 48n = 636 = 0
⇒ n(4n + 53) - 12(4n + 53) = 0
⇒ (4n + 53) (n - 12) = 0
⇒ 4n + 53 = 0 or n - 12 = 0
⇒ n = `(-53)/4` or n = 12
As the number of terms can neither be negative nor fractional, therefore, n = 12 only.
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