Let S_n = 12+16+112+120+ up to n terms.

Question

Let $S_n = \dfrac12+\dfrac16+\dfrac{1}{12}+\dfrac{1}{20}+\cdots$ up to $n$ terms. If the sum of the first six terms of an A.P. with first term $-p$ and common difference $p$ equals $\sqrt{2026\cdot S_{2025}}$, then the absolute difference between the $20$th and $15$th terms of the A.P. is:

Accepted Answer

$S_n = \displaystyle\sum_{k=1}^{n}\frac{1}{k(k+1)} = 1-\frac{1}{n+1} = \frac{n}{n+1}$, so $S_{2025}=\dfrac{2025}{2026}$.

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

For the A.P. with first term $-p$, common difference $p$: $S_6 = \dfrac62[2(-p)+5p] = 3(3p)=9p = 45 \Rightarrow p=5.$

$$|A_{20}-A_{15}| = |(-p+19p)-(-p+14p)| = |18p-13p| = 5p = 25.$$

Solutions · 1