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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?

Answers (1)

best_answer

Given that 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
 

Posted by

Gautam harsolia

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