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

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#### Solution

Let there be *n* terms of this A.P.

For this A.P., *a* = 9

*d* = *a*_{2} − *a*_{1} = 17 − 9 = 8

`S_n = n/2[2a+(n-1)d]`

`636 = n/2[2xxa+(n-1)8]`

`636 = n/2[18+(n-1)8]`

636 = *n *[9 + 4*n* − 4]

636 = *n *(4*n* + 5)

4*n*^{2} + 5*n* − 636 = 0

4*n*^{2} + 53*n* − 48*n* − 636 = 0

*n *(4*n* + 53) − 12 (4*n* + 53) = 0

(4*n* + 53) (*n* − 12) = 0

Either 4*n *+ 53 = 0 or *n* − 12 = 0

`n = (-53)/4` or n = 2

*n* cannot be `(-53)/4`

As the number of terms can neither be negative nor fractional, therefore, *n* = 12 only.

Concept: Sum of First n Terms of an AP

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