Sequence and series JEE Main 2024 with Solution

This Sequence and series JEE Main 2024 with Solution explained here with detailed explanation. This question helps student understand Arithmetic Progression(AP), Geometric Progression(GP).

Question :

If log_e a , log_e b, log_e c are in A.P. and log_e a – log_e 2b, log_e 2b – log_e 3c, log_e 3c – log_e a are also in A.P then a : b : c is equal to

Options :

(A) 6 : 3 : 2

(B) 9 : 6 : 4

(C) 25 : 10 : 4

(D) 16 : 4 : 1

Sequence and series JEE Main 2024 with Solution

Solution :

Given

log_e a, log_e b, log_e c are in A.P

log_e b – log_e a = log_e c – log_e b

log_e b + log_e b = log_e c + log_e a

2log_e b = log ac

log_e b² = log ac

b² = ac

log_e a – log_e 2b, log_e 2b – log_e 3c, log_e 3c – log_e a

log [$\frac{a}{2b}$], log [$\frac{2b}{3c}$], log [ $\frac{3c}{a}$ ]

2log [ $\frac{2b}{3c}$ ] = log [ $\frac{a}{2b}$ ] + log [ $\frac{3c}{a}$ ]

log [ $\frac{2b}{3c}$ ]² = log [ $\frac{a3c}{2ba}$ ]

(2b)³ = (3c)³

2b = 3c

b = $\frac{3c}{2}$

$\frac{b}{c}$ = $\frac{3}{2}$

b² = ac

b . b = ac

$\frac{b}{a}$ = $\frac{a}{b}$ = $\frac{3}{2}$

$\frac{b}{c}$ = $\frac{3}{2}$ x $\frac{2}{2}$

$\frac{b}{c}$ = $\frac{6}{4}$

$\frac{a}{b}$ = $\frac{3}{2}$ x $\frac{3}{3}$

$\frac{a}{b}$ = $\frac{9}{6}$

a : b : c = 9 : 6 : 4

Correct answer : Option (B)

Practice Questions :

1. Insert three numbers between 1 and 256 so that the resulting sequence is a G.P

solution :

1, G₁, G₂, G₃, 256

a = 1

ar⁴ = 256

(1)r⁴ = 256

r = $\pm$4

If r = +4

G₁ = ar = (1)(4) = 4

G₂ = ar² = (1)(4)² = 16

G₃ = ar³ = (1)(4)³ = 64

If r = -4

G₁ = ar = (1)(-4) = -4

G₂ = ar² = (1)(-4)² = 16

G₃ = ar³ = (1)(-4)³ = -64

The resulting sequence is 4, 16, 64

2. If A.M and G.M of two positive numbers a and b 11are 10 and 8 respectively, find the numbers

Solution :

A.M = 10

$\frac{a+b}{2}$ = 10

a + b = 20 —————————- 1

G.M = 8

($\sqrt{ab^2}$) = (8)²

ab = 64 —————————-2

we know that

(a + b)² – (a -b)² = 4ab

(a – b)² = (a + b)² – 4ab

(a – b)² = (20)² – 4(64)

= 400 – 256

(a – b)² = 144 $\rightarrow$ a – b = $\pm\sqrt{144}$

a – b = $\pm$ 12————————- 3

add equation 1 + 3

a + b + a – b = 20 + 12

2a = 32

a = $\frac{32}{2}$

a = 16

substitude a = 16 in equation 1

a + b = 20

b = 20 – 16

b = 4

a + b + a – b = 20 – 12

2a = 8

a = 4

sub a = 4 in equation 1

a + b = 20

b = 20 – 4

b = 16

The two numbers are 4, 16 (or) 16, 4

3. Find the 12^th term of a G.P whose 8^th term is 192 and the common ratio is 2

Solution :

a₈ = 192 , r = 2

ar⁷ = 192

a. 2⁷ = 192

a = $\frac{192}{128}$

a = $\frac{3}{2}$

a₁₂ = ar^{12 – 1}

= ($\frac{3}{2}$)(2)¹¹

= 3 x 2¹⁰

= 3 x 1024

= 3072

4. The 5^th, 8^th and 11^th terms of a G.P are p, q and s respectively show that q² = ps

Solution :

5^th term = ar⁴ = p

8^th term = ar⁷ = q

11^th term = ar¹⁰ = s

LHS = q²

= (ar⁷)²

= a². r¹⁴

RHS = p.s

= (ar⁴)(ar¹⁰)

= a².r^{4 + 10}

= a².r¹⁴

Here LHS = RHS q² = ps Hence proved

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