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How many terms of the AP $9, 17, 25, \ldots$ must be taken to give a sum of $636$?
A10
B11
C12
D13
Answer & Solution
Correct answer: C. 12
Here $a=9$ and $d=8$. Using $S_n=\frac{n}{2}[2a+(n-1)d]$, we get
$$636=\frac{n}{2}[18+8(n-1)]=\frac{n}{2}(8n+10)=n(4n+5).$$
So $4n^2+5n-636=0$. Solving gives $n=12$ (the other root is negative, so it is rejected).
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