Let S_n=(1)/(2)+(1)/(6)+(1)/(12)+(1)/(20)+ upto n terms.

Question

Let $S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\ldots$ upto $n$ terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is $\sqrt{2026 \mathrm{~S}_{2025}}$, then the absolute difference bet...

Accepted Answer

To find the value of ${S}_{2025}$, calculate the partial sum

${S}_{n}=\sum _{n=1}^{2025}\frac{1}{n(n+1)}=\sum _{n=1}^{2025}(\frac{1}{n}-\frac{1}{n+1})$

This series is telescopic, meaning most terms cancel out with each other:

$=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots +(\frac{1}{2025}-\frac{1}{2026})$

Simplifying the expression, we find:

$=\frac{1}{1}-\frac{1}{2026}=1-\frac{1}{2026}=\frac{2025}{2026}$

Now, evaluate $\sqrt{2026\cdot {S}_{2025}}$:

$\sqrt{2026\cdot \frac{2025}{2026}}=\sqrt{2025}=45$

For the arithmetic progression with the first term $-p$ and common difference $p$, the sum of the first six terms is given by:

$\frac{6}{2}[-2p+(6-1)p]=3(5p)=15p$

Given that:

$15p=45$

We find:

$p=3$

To determine the absolute difference between the ${20}^{}$ and ${15}^{}$ terms of the A.P., compute:

$|{A}_{20}-{A}_{15}|=|(-p+19p)-(-p+14p)|$

Substituting the value of $p$:

$=|18p-13p|=|5p|=5	imes 5=25$

Thus, the absolute difference is $25$.

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