Sequences & Series
Telescoping Sum Combined with AP
nta_pyq_2025_apr
Grade 11
Question:
Let $S_n = \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdots$ to $n$ terms. If the sum of the first 6 terms of an AP with first term $-p$ and common difference $p$ equals $\sqrt{2026\cdot S_{2025}}$, and an denotes the $n^{\text{th}}$ term of this AP, then $|A_{20}-A_{15}|$ equals
Step-by-Step Solution
Key Concept: Telescope $S_n=\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}$, compute $\sqrt{2026\cdot S_{2025}}=45$, then find $p$ from the AP sum and use $A_n=(n-2)p$.
$S_n=\displaystyle\sum_{k=1}^n\dfrac{1}{k(k+1)}=\dfrac{n}{n+1}$, so $S_{2025}=\dfrac{2025}{2026}$.
$\sqrt{2026\cdot S_{2025}}=\sqrt{2026\cdot\dfrac{2025}{2026}}=\sqrt{2025}=45$.
Sum of first 6 AP terms $= 6(-p)+\dfrac{6\cdot5}{2}\cdot p=-6p+15p=9p=45\Rightarrow p=5$.
$A_n=-p+(n-1)p=(n-2)p$.
$|A_{20}-A_{15}|=|(20-2)p-(15-2)p|=|18p-13p|=5p=25$.
Correct Answer: 4