Let T_ r be the r^ th term of an A.P. If for some m, T_ m =(1)/(25), T_ 25 =(1)/(20) , and 20 ∑_ r=1 ^ 25 T_ r =13 , then 5 m ∑_ r=m ^ 2 m T_ r is equal to

Correct answer: B. 126 aligned & T_ m =(1)/(25), ~T_ 25 =(1)/(20), 20 ∑_ r=1 ^ 25 ~T_ r =13 & ~T_ m =a+(m-1) d=(1)/(25) .(1) & T_ 25 =a+24 ~d=(1)/(20) & 20 · (25)/(2)[a+(1)/(20)]=13 a=(1)/(500) aligned also, 20 ~S_ 25 =20 · (25)/(2)[2 a+24 ~d]=13 ~d=(1)/(500) from (1) (1)/(500)+ m-1 500 =(1)/(25) ~m=20 Now, 5 ~m ∑_ r=m ^ 2 ~m ~T_ r =100 ∑_ r=20 ^ 40 ~T_ r =126

JEE Main 202528 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 28 January · Morning Shift · Mathematics · Sequences and Series

Let T_ r be the r^ th term of an A.P. If for some m, T_ m...

Let

\mathrm{T}_{\mathrm{r}}

be the

\mathrm{r}^{\text{th }}

term of an A.P. If for some

\mathrm{m}, \mathrm{T}_{\mathrm{m}}=\frac{1}{25}, \mathrm{ T}_{25}=\frac{1}{20}

, and

20 \sum_{\mathrm{r}=1}^{25} \mathrm{ T}_{\mathrm{r}}=13

, then

5 \mathrm{ m} \sum_{

\mathrm{r}=\mathrm{m}

^{2 \mathrm{ m}} \mathrm{ T}_{\mathrm{r}}

is equal to

Options

A98
B126
C142
D112

Correct answer

B. 126

Step-by-step solution

\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}=20

Now,

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

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