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If the sum of the first $7$ terms of an AP is $49$ and that of the first $17$ terms is $289$, what is the sum of the first $n$ terms?
A$S_n=n^2$
B$S_n=n(2n-1)$
C$S_n=n(n+1)$
D$S_n=n^2+n$
Answer & Solution
Correct answer: A. $S_n=n^2$
For an AP, $S_n=\frac{n}{2}[2a+(n-1)d]$. Given
$$\frac{7}{2}(2a+6d)=49 \Rightarrow a+3d=7$$
and
$$\frac{17}{2}(2a+16d)=289 \Rightarrow a+8d=17.$$
Subtracting gives $5d=10$, so $d=2$. Then $a+6=7$, hence $a=1$. Therefore
$$S_n=\frac{n}{2}[2+(n-1)2]=\frac{n}{2}(2n)=n^2.$$
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