Problem 638

Sir SagarMarch 18, 2015March 18, 20150

Find S25 for $\frac{1}{2}+1+\frac{3}{2}+2+...$

Solution:-

We know S = $\frac{n}{2}[2a_{1}+(n-1)d]$ , Where a1 = first term , n = finding term , d = second term – first term

S25 = $\frac{25}{2}[2*(\frac{1}{2})+(25-1)(\frac{1}{2})]$

S25 = $\frac{325}{2}$

So the value of S25 $\frac{325}{2}$

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