Question

# The houses of a row are numbered consecutively from $$1$$ to $$49$$. Show that there is a value of $$x$$ such that the sum of the numbers of the houses preceding the house numbered $$x$$ is equal to the sum of the numbers of the houses following it. Find this value of $$x$$.

Solution

## Here $$a=1, d=1$$ and $$a_{49}=49$$As per question, $$S_{x-1}=S_{49}-S_{x}$$............(i)S_{x}=\dfrac{x}{2}\left [ 2+(x-1)\times 1 \right ]=\dfrac{x}{2}(x+1)=\dfrac{x^{2}+x}{2}$$Similarly,$$S_{x-1}=\dfrac{x-1}{2}\left [ 2+(x-1-1)\times 1 \right ]=\dfrac{x-1}{2}(2+x-2)=\dfrac{(x-1)x}{x}\dfrac{x^{2}-x}{2}$$Similarly,$$S_{49}=\dfrac{49}{2}\left[ 2+48\times 1 \right ]=\dfrac{49}{2}\times 50=1225$$After substituting the values of$$S_{x-1}, S_{49}$$and$$S_x$$in equation (1), we get$$S_{x-1}==S_{49}-S_{x}\dfrac{x^{2}-x}{2}=1225-\dfrac{x^{2}+x}{2}\dfrac{x^{2}-x}{2}+\dfrac{x^{2}+x}{2}=1225\dfrac{x^{2}-x+x^{2}-x}{2}=1225x^{2}=1225x=35MathematicsRS AgarwalStandard X

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