Sequence and series JEE Main 2025 PYQ with solution

Question :

Solution :

Given the quadratic equation is 3x²– px + q = 0

roots are 10^th and 11^th terms

Sum of roots = T₁₀ + T₁₁ = $\frac{p}{3}$

a + 9d + a + 10d = $\frac{p}{3}$

2a + 19d = $\frac{p}{3}$

Given d = $\frac{3}{2}$

2a + 19.$\frac{3}{2}$ = $\frac{p}{3}$

2a + $\frac{57}{2}$ = $\frac{p}{3}$ ——————————————-1

Product of root = ( T₁₀ )( T₁₁ ) = $\frac{q}{3}$

( a + 9d )( a + 10d) = $\frac{q}{3}$ ———————————-2

Sum of 11 terms = 88

$\frac{11}{2}$(2a + 10d ) = 88

a + 5d = 8 ( Substitute d = $\frac{3}{2}$ )

a + $\frac{15}{2}$ = 8

a = 8 – $\frac{15}{2}$ = $\frac{1}{2}$

Substitute a = $\frac{1}{2}$ in 1^st equation

2( $\frac{1}{2}$ ) + 57 = $\frac{p}{3}$

1 + $\frac{57}{2}$ = $\frac{p}{3}$

$\frac{59}{2}$ = $\frac{p}{3}$

2p = 177

Substitute a = $\frac{1}{2}$, d = $\frac{3}{2}$

( $\frac{1}{2}$ + 9( $\frac{3}{2}$ ) ($\frac{1}{2}$ + 10($\frac{3}{2}$) = $\frac{q}{3}$

($\frac{1}{2}$ + $\frac{27}{2}$ ) ($\frac{1}{2}$ + $\frac{30}{2}$ ) = $\frac{q}{3}$

( 14) ($\frac{31}{2}$) = $\frac{q}{3}$

7 (31)(3) = q

q = 651

q – 2p = 651 – 177 = 474

Formula used :

S_n = $\frac{n}{2}$ [ 2a + (n – 1)d ]

Sum of terms – –$\frac{b}{a}$

Product of terms – $\frac{c}{a}$

Practice Questions :

1. If the first term of the an AP is -1 and Common difference is – 3 then find the 21^st term

Solution :

a = -1

d = -3

n = 21

T₂₁ = -1 + ( 21 – 1)d = – 1 + (20) -3 = -1-60 = -61

Formula used = a_n = a + ( n – 1 )d

2. Which term of the AP 101, 99, 97,…………………………….. is 47 ?

Solution :

101, 99, 97, 95,…………………………….47,45,43,…….

a = 101 d = -2

T_n = a + (n – 1)d

47 = 101 + ( n -1 ) -2

47 = 101 – 2n + 2

47 – 103 = -2n

-56 = – 2n

n = 28

The n^th term = 28

3. Find the 16^th term from the end of the AP 2 , 6, 10………………………….86 ?

Solution :

86, 82, 78, ……………………8, 6, 2

a = 86 d = 4

T_m = a + (m – 1)-d

T₁₆ = 86 + ( 16 – 1)-4

86 + (15)-4

T₁₆ = 26

4. Find the sum of the First 29 terms of the AP 2, 5, 8, 11………………………….

Solution :

S_n = $\frac{n}{2}$ ( 2a + (n -1)d

a = 2 d = 3 n = 29

S₂₉ = $\frac{29}{2}$[ 2(2) + (28)3 ]

29 ($\frac{4 + (28)3}{2}$) = 29 ( 2 + (14)3 ) = 29 (44) = 1276

5. If for a sequence S_n = 5n² + 3n then find the n^th term of the sequence

Solution :

Method – 1

S_n = 5n² + 3n

d = 2p = 2 x 5 = 10

a = p + q = 8

T_n = a + (n -1)d

T_n = 10n – 2

Method – 2

S_n = n ( 5n +3) = n( 5(n – 1) +3 )

= n ( 16 + (n – 1)10 )

a = 8 d = 10

T_n = a + (n – 1)d = 8 + (n -1) 10 = 10n – 2

6. Find 10^th Common term in 3, 7, 11, 15, 19 ……………. and 1, 6, 11, 16 , 21………………

Solution :

Common term in both series : 11

3, 7, 11, 15………………AP

d₁ = 4

1, 6, 11, 16………………….

d₂ = 5

d = LCM ( d₁ , d₂ )

= LCM [ 4, 5]

d = 20

11, 31, 51…………………

a + 9d = 11 + 9 (20) = 11 + 180 = 191

7. If the sum of three numbers which are in AP is 27 and product of first and last is 77, then Find the numbers

Solution :

a – d , a , a +d

sum of terms = a – d + a + a + d = 27

3a = 27

a = 9

Product of terms = ( a – d )(a + d) = 77

a² – d² = 77

81 – d² = 77

81 – 77 = d²

d² = 4

a = 9 d = 2

a – d , a , a + d = 9 – 2, 9, 9 + 2 = 7, 9, 11

a = 9 d = -2

9 – (-2) , 9, 9 – 2 = 11, 9, 7

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