Sequence and series JEE Main 2025 shift – I with solution

Question :

Let $T_r$be the r^th term of an AP If for some m. T_m = $\frac{1}{25}$, T₂₅ = $\frac{1}{20}$ and 20$\sum_{r=1}^{25}T_r$ = 13, then 5m$\sum_{r=m}^{2m}T_r$ is

Options :

(A) 98

(B) 126

(C) 112

(D) 142

Sequence and series JEE Main 2025 shift – I with solution

Answer :

Given

20$\sum_{r=1}^{25}T_r$ = 13

20 [ T₁ + T₂ + T₃ + T₄ ………………+ T₂₅ ] = 13

use S_n formula

S_n = $\frac{n}{2}$[ a + l ]

20$\frac{25}{2}$[ T₁ + T₂₅ ] = 13

250 [ a + $\frac{1}{20}$ ] = 13

250a + $\frac{25}{2}$ = 13

250a = 13 – $\frac{25}{2}$

250a = $\frac{1}{2}$

a = $\frac{1}{500}$

T₂₅ = $\frac{1}{20}$ [ Given in question ]

T₂₅ = $\frac{1}{500}$ + 24d

$\frac{1}{20}$ = $\frac{1}{500}$ + 24d

24d = $\frac{1}{20}$ – $\frac{1}{500}$

24d = $\frac{25 – 1}{500}$

24d = $\frac{24}{500}$

d = $\frac{1}{500}$

T_m = a + ( m -1)d

$\frac{1}{25}$ = $\frac{1}{500}$ + ( m -1)$\frac{1}{500}$

$\frac{1}{25}$ = $\frac{m}{500}$

m = 20

5m$\sum_{r=m}^{2m} T_r$

5 x 20 ( T₂₀ + T₂₁ + T₂₂ + T₂₃ + …………………….+ T₄₀ )

100 [ a + 19d + a + 20d + ……………… + a + 39d ]

100 [ 21a + d (19 + 20 + ………….. + 39 ]

100 [ 21a + d$\frac{21}{2}$ [ 19 + 39 ]

(a) (21) (100) [ 1 + $\frac{58}{2}$ ]

21 x $\frac{1}{500}$ x 100 [ 1 + 29 ]

$\frac{21}{500}$ x 100 x 30

21 x 6 = 126

Correct answer : Option (B)