Sequence and series JEE Main 2025 – Important PYQ

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 – Important PYQ

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)

Practice Questions :

Find the sum to infinity in each of the following Geometric Progression

1. 1, $\frac{1}{3}$, $\frac{1}{9}$ …………………

Solution :

S_$\infin$ = $\frac{a}{1 – r}$

= $\frac{1}{1 – 1/3}$ = $\frac{1}{3 – 1/3}$ = $\frac{1}{2/3}$ = $\frac{3}{2}$ = 1.5

2. 6, 1.2, 2.4 , …………………..

Solution :

a = 6, r = $\frac{1.2}{6}$ = $\frac{12}{6 x 10}$ = $\frac{1}{5}$

s_$\infin$ = $\frac{a}{1 – r}$ = $\frac{6}{1 – \frac{1}{5}}$ = $\frac{6}{\frac{5 – 1}{5}}$ = $\frac{6 x 5}{4}$ = $\frac{15}{2}$ = 7.5

3. 5, $\frac{20}{7}$, $\frac{80}{49}$, …………………………….

Solution :

a = 5 , r = $\frac{20/7}{5/1}$ = $\frac{20}{7}$ x $\frac{1}{5}$ = $\frac{4}{7}$

S_$\infin$ = $\frac{a}{1 – r}$ = $\frac{5}{1 – \frac{4}{7}}$ = $\frac{5 x 7}{3}$ = $\frac{35}{3}$

4. $\frac{-3}{4}$, $\frac{3}{16}$, $\frac{-3}{64}$, ………………..

Solution :

a = $\frac{-3}{4}$, r = $\frac{3/16}{-3/4}$ = $\frac{-1}{4}$

s_$\infin$ = $\frac{a}{1 – r }$ = $\frac{-3/4}{ 1 + \frac{1}{4}}$ = $\frac{-3}{5}$

5. The sum of first three terms of a G.P is $\frac{39}{40}$ and their product is 1 find the common ratio and the terms

Solution :

Let the first three terms be $\frac{a}{r}$, a, ar

Given

Sum of terms : $\frac{a}{r}$ + a + ar = $\frac{39}{10}$ $\rightarrow$ 1

Product of terms : $\frac{a}{r}$.a.ar = 1 $\rightarrow$ a³ = 1

a = 1

Sun a = 1 in eq 1

$\frac{1}{r}$ + 1 + r = $\frac{39}{10}$

$\frac{1 + r + r^2}{r}$ = $\frac{39}{10}$

10 + 10r + 10r² = $\frac{39}{10}$

5r[ 2r – 5] -2[ 2r – 5] = 0

(2r – 5) ( 5r – 2) = 0

2r – 5 = 0 (or) 5r = 2

2r = 5 (or) r = $\frac{2}{5}$

r = $\frac{5}{2}$

If a = 1, r = $\frac{5}{2}$

The three terms are $\frac{2}{5}$, 1, $\frac{5}{2}$

If a = 1, r = $\frac{2}{5}$

The three terms are $\frac{5}{2}$, 1, $\frac{2}{5}$

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