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 $$a=1$$$$S_{n}=\dfrac{n}{2}\left [ 2(1)+(n-1)2 \right ]$$$$=\dfrac{n}{2}\left [ 2+(n-1)2 \right ]$$$$=\dfrac{n}{2}(2+2n-2)=n^{2}$$MathematicsRS AgarwalStandard X

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