Sequence and series JEE Main 2024 with solution

Question :

In an increasing geometric progression of positive session terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is 49 Then the sum of the 4^th, 6^th and 8^th terms is

Options :

(A) 96

(B) 78

(C) 91

(D) 84

Sequence and series JEE Main 2024 with solution

Solution :

Given

T₂ + T₆ = $\frac{70}{3}$

ar + ar⁵ = $\frac{70}{3}$

ar ( 1 + r⁴ ) = $\frac{70}{3}$ ————————— 1

T₃ . T₅ = 49

ar² . ar⁴ = 49

a². r⁶ = 7²

( ar³ )² = 7²

ar³ = 7 ————————————- 2

Divide equation $\frac{1}{2}$

$\frac{ar ( 1 + r^4)}{ar^3}$ = $\frac{70/3}{7}$

$\frac{1 + r^4}{r^2}$ = $\frac{10}{3}$

3 + 3r⁴ – 10r² = 0

3( r²)² – 10r² + 3 = 0

3(r²)² – 9r² – r² + 3 = 0

3r² [ r² – 3 ] – 1[ r² -3 ] = 0

[ r² – 3 ] [ 3r² – 1 ] = 0

r² = 3 or 3r² = 1

r = $\sqrt{3}$ or r² = $\frac{1}{3}$

T₄ + T₆ + T₈ = ar³ + ar⁵ + ar⁷

= ar³ [ 1 + r² + r⁴ ]

= 7 [ 1 + 3 + 9 ]

= 7 ( 13 )

= 91

Correct answer : Option (C)

Practice Questions :

1. Write the first five terms of each of the sequence in exercise 1 to 6 whose n^th terms are :

i) a_n = n ( n + 2)

Solution :

a₁ = 1 ( 1 + 2 ) = 1( 3 ) = 3

a₂ = 2 ( 2 + 2 ) = 2( 4 ) = 8

a₃ = 3 ( 3 + 2) = 3( 5 ) = 15

a₄ = 4 ( 4 + 2 ) = 4( 6 ) = 24

a₅ = 5( 5 + 2) = 5( 7 ) = 35

• The first five terms are 3, 8 , 15, 24 , 35

ii) a_n = $\frac{n}{n + 1}$

Solution :

a₁ = $\frac{1}{1 + 1}$ = $\frac{1}{2}$

a₂ = $\frac{2}{2 + 1}$ = $\frac{2}{3}$

a₃ = $\frac{3}{3 + 1}$ = $\frac{3}{4}$

a₄ = $\frac{4}{4 +1}$ = $\frac{4}{5}$

a₅ = $\frac{5}{5 + 1}$ = $\frac{5}{6}$

The first five terms are $\frac{1}{2}$, $\frac{2}{3}$, $\frac{3}{4}$, $\frac{4}{5}$, $\frac{5}{6}$

iii) a_n = 2ⁿ

a₁ = 2¹ = 2

a₂ = 2² = 4

a₃ = 2³ = 8

a₄ = 2⁴ = 16

a₅ = 2⁵ = 32

The first five terms are 2, 4, 8, 16, 32

iv) a_n = $\frac{2n – 3}{6}$

solution :

a₁ = $\frac{2(1) – 3 }{6}$ = $\frac{ 2 – 3}{6}$ = $\frac{-1}{6}$

a₂ = $\frac{ 2(2) – 3}{6}$ = $\frac{4 – 3}{6}$ = $\frac{1}{6}$

a₃ = $\frac{2(3) – 3}{6}$ = $\frac{6 – 3}{6}$ = $\frac{3}{6}$ = $\frac{1}{2}$

a₄ = $\frac{2(4) -3}{6}$ = $\frac{8 – 3}{6}$ = $\frac{5}{6}$

a₅ = $\frac{2(5) – 3}{6}$ = $\frac{10 – 3}{6}$ = $\frac{7}{6}$

The first five terms are $\frac{ – 1}{6}$, $\frac{1}{6}$, $\frac{1}{2}$, $\frac{5}{6}$, $\frac{7}{6}$

v) a_n = ( – 1)^{n – 1} 5^{n + 1}

solution :

a₁ = ( -1)^{1 – 1} 5^{1 + 1} = (-1)⁰. 5² = 1 x 25 = 25

a₂ = ( -1)^{2 -1} 5^{2 + 1} = ( -1)¹. 5³ = -125

a₃ = ( -1)^{3 – 1}. 5^{3 + 1} = ( -1)². 5⁴ = 625

a₄ = ( -1)^{4 – 1}. 5^{4 + 1} = ( – 1)³. 5⁵ = – 3125

a₅ = ( -1)^{5 – 1}. 5^{5 + 1} = ( – 1)⁴. 5⁶ = 15625

The first five terms are 25, -125, 625, – 3125, 15625

vi) a_n = n. $\frac{n^2 +5}{4}$

solution :

a₁ = 1 x $\frac{1^2 + 5}{4}$ = 1 x $\frac{1 + 5}{4}$ = 1 x $\frac{6}{9}$ = $\frac{3}{2}$

a₂ = 2 x $\frac{2^2 + 5 }{4}$ = 2 x $\frac{ 4 + 5}{4}$ = $\frac{9}{2}$

a₃ = 3 x $\frac{3^2 + 5 }{4}$ = 3 x $\frac{9 + 5}{4}$ = 3. $\frac{14}{4}$ = $\frac{21}{2}$

a₄ = 4 x $\frac{4^2 + 5}{4}$ = 4 x $\frac{ 16 + 5 }{4}$ = 21

a₅ = 5 x $\frac{5^2 + 5}{4}$ = 5. $\frac{25 + 5}{4}$ = 5. $\frac{30}{4}$ = $\frac{75}{2}$

The first five terms are $\frac{3}{2},$$\frac{9}{2}$, $\frac{21}{2}$, 21, $\frac{75}{2}$

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