Sequences & Series
Sequences and Series
nta_pyq_2025_jan
Grade 11
Question:
Let $T_{r}$ be the $r^{\text{th}}$ term of an A.P. If for some $m$, $T_{m}=\dfrac{1}{25}$ and $T_{25}=\dfrac{1}{20}$, and $\displaystyle\sum_{r=1}^{25}T_{r}=\dfrac{13}{20}$, then $5m\,\displaystyle\sum_{r=m}^{2m}T_{r}$ is equal to:
Step-by-Step Solution
Key Concept: Three equations $T_{m}=\tfrac{1}{25}$, $T_{25}=\tfrac{1}{20}$, $S_{25}=\tfrac{13}{20}$ fix $a$, $d$, $m$. The clean form $T_{r}=r/500$ then makes the partial sum a triangle-number computation.
$S_{25}=\dfrac{25}{2}(2a+24d)=\dfrac{13}{20}\Rightarrow a+12d=\dfrac{13}{500}.$
$T_{25}=a+24d=\dfrac{1}{20}=\dfrac{25}{500}.$
Subtract: $12d=\dfrac{12}{500}\Rightarrow d=\dfrac{1}{500}$, so $a=\dfrac{1}{500}$ and $T_{r}=\dfrac{r}{500}.$
$T_{m}=\dfrac{m}{500}=\dfrac{1}{25}=\dfrac{20}{500}\Rightarrow m=20.$
$$5m\sum_{r=20}^{40}T_{r}=100\sum_{r=20}^{40}\frac{r}{500}=\frac{1}{5}\cdot\frac{(20+40)(40-20+1)}{2}=\frac{1}{5}\cdot\frac{60\cdot 21}{2}=\frac{1260}{10}=126.$$
Correct Answer: 2