Sequences & Series
AP with Three Given Conditions
nta_pyq_2025_apr
Grade 11
Question:
Let $T_r$ be the $r^{\text{th}}$ term of an AP. If $T_m=\dfrac{1}{25}$, $T_{25}=\dfrac{1}{20}$, and $20\displaystyle\sum_{r=1}^{25}T_r=13$, then $5m\displaystyle\sum_{r=m}^{2m}T_r$ equals
Step-by-Step Solution
Key Concept: Use the three given conditions to determine $a$, $d$, and $m$ uniquely, then note that $\sum_{r=m}^{2m}$ has $m+1$ terms and use the AP sum formula.
From $20\cdot\sum_{r=1}^{25}T_r=13$: $\sum_{r=1}^{25}T_r=\frac{13}{20}=\frac{25}{2}(2a+24d)\Rightarrow a+12d=\frac{13}{500}$.
From $T_{25}=a+24d=\frac{1}{20}$, subtracting: $12d=\frac{1}{20}-\frac{13}{500}=\frac{25-13}{500}=\frac{12}{500}\Rightarrow d=\frac{1}{500}$, $a=\frac{1}{500}$.
$T_m=a+(m-1)d=\frac{m}{500}=\frac{1}{25}\Rightarrow m=20$.
$5m=100$. $\sum_{r=20}^{40}T_r=\frac{21}{2}(T_{20}+T_{40})=\frac{21}{2}\left(\frac{20}{500}+\frac{40}{500}\right)=\frac{21}{2}\cdot\frac{60}{500}=\frac{63}{50}$.
$5m\sum_{r=m}^{2m}T_r=100\cdot\frac{63}{50}=126$.
Correct Answer: 2