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Question

If the sum of first $$7$$ terms of an AP is $$49$$ and that of $$17$$ terms is $$289$$, find the sum of first $$n$$ terms.


Solution

Given 
$$S_{7}=49$$ and $$S_{17}=289$$

By using $$S_{n}=\dfrac{n}{2}\left [ 2a+(n-1)d \right ]$$ we have,


$$S_{7}=\dfrac{7}{2}\left [ 2a+(7-1)d \right ]=49$$

$$\Rightarrow 49=\dfrac{7}{2}\left [ 2a+(7-1)d \right ]$$

$$\Rightarrow 49=\dfrac{7}{2}(2a+6d)$$

$$\Rightarrow 7=a+3d$$

$$\Rightarrow a+3d=7$$...................(i)

$$S_{17}=\dfrac{17}{2}\left [ 2a+(17-1)d \right ]=289$$

$$\Rightarrow 289=\dfrac{17}{2}\left [ 2a+(17-1)d \right ]$$

$$\Rightarrow 289=\dfrac{17}{2}(2a+16d)$$

$$\Rightarrow 17=a+8d$$

$$\Rightarrow a+8d=17$$......................(ii)

Substituting (i) from (ii), we get

$$5d= 10$$ or $$d=2$$

From equation (i),

$$a+3(2)=7$$

$$a+6=7$$ or $$a=1$$

$$S_{n}=\dfrac{n}{2}\left [ 2(1)+(n-1)2 \right ]$$

$$=\dfrac{n}{2}\left [ 2+(n-1)2 \right ]$$

$$=\dfrac{n}{2}(2+2n-2)=n^{2}$$

Mathematics
RS Agarwal
Standard X

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