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

Q : 4     How many terms of the AP :  $\small 9,17,25,...$ must be taken to give a sum of $\small 636$?

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Gievn AP is
$\small 9,17,25,...$
Here, $a =9 \ and \ d = 8$
And  $S_n = 636$
Now , we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 636 = \frac{n}{2}\left \{ 18+(n-1)8 \right \}$
$\Rightarrow 1272 = n\left \{ 10+8n \right \}$
$\Rightarrow 8n^2+10n-1272=0$
$\Rightarrow 2(4n^2+5n-636)=0$
$\Rightarrow 4n^2+53n-48n-636=0$
$\Rightarrow (n-12)(4n+53)=0$
$\Rightarrow n = 12 \ \ and \ \ n = - \frac{53}{4}$
Value of  n  can not be negative so only value of n is 12
Therefore, Sum of 12 terms of AP  $\small 9,17,25,...$ must be taken to give a sum of $\small 636$

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