$\begin{aligned}
& \mathrm{T}{\mathrm{m}}=\frac{1}{25}, \mathrm{~T}{25}=\frac{1}{20}, 20 \sum_{\mathrm{r}=1}^{25} \mathrm{T}_{\mathrm{r}}=13 \
& \mathrm{T}{\mathrm{m}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=\frac{1}{25} \ldots \ldots .(1) \
& \mathrm{T}{25}=\mathrm{a}+24 \mathrm{d}=\frac{1}{20} \
& 20 \cdot \frac{25}{2}\left[\mathrm{a}+\frac{1}{20}\right]=13 \Rightarrow \mathrm{a}=\frac{1}{500}
\end{aligned}also,20 \mathrm{S}{25}=20 \cdot \frac{25}{2}[2 \mathrm{a}+24 \mathrm{d}]=13 \Rightarrow \mathrm{d}=\frac{1}{500}from(1)\frac{1}{500}+\frac{\mathrm{m}-1}{500}=\frac{1}{25} \Rightarrow \mathrm{m}=20Now,5 \mathrm{m} \sum{\mathrm{r}=\mathrm{m}}^{2 \mathrm{m}} \mathrm{T}{\mathrm{r}}=100 \sum{\mathrm{r}=20}^{40} \mathrm{~T}_{\mathrm{r}}=126$
Let T_ be the ^ th term of an A.P. If for some m, T_ m= 1 25, ~T_25= 1 20, and 20 Σ_ =1^25 ~T_ =13, then 5 ~m Σ_ = m^2 ~m ~T_ is equal to
JEE Main 2025 — Mathematics Algebra
Let Tr be the rth term of an A.P. If for some m,Tm=251, T25=201, and 20r=1∑25 Tr=13, then 5 mr=m∑2 m Tr is equal to
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
98
- B
126
- C
142
- D
112
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