JEE Main 2025 Sequence and series PYQ with solution

Question :

In an Arithmetic Progression , If S₄₀ = 1030 and S₁₂ = 57, then S₃₀ – S₁₀ is equal to

Options :

(a) 505

(b) 515

(c) 510

(d) 525

JEE Main 2025 Sequence and series PYQ with solution

Answer :

Step : 1 Given,

S₄₀ = 1030

$\frac{40}{2}$(2a + 39d ) = 1030

40a + 780d = 1030

4a + 78d = 103 ——————————1

S₁₂ = 57

$\frac{12}{2}$(2a + 11d ) = 57

6( 2a + 11d ) = 57

4a + 22d = 19————————————-2

Step : 2

Solving 1 and 2 equation

4a + 78d – 4a – 22d = 103 – 19

56d = 84

d = $\frac{84}{56}$

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

Step : 3

substitute d = $\frac{3}{2}$ in 2nd equation

4a + 33 = 19

4a = 19 – 33

a = –$\frac{14}{4}$

a = –$\frac{7}{2}$

Step : 4

S₃₀ – S₁₀ = $\frac{30}{2}$( 2a + 29d ) – $\frac{10}{2}$( 2a + 9d )

= 30a + 435d – 10a – 45d

= 20a + 390d

Step : 5

substitute a = –$\frac{7}{2}$ , d = $\frac{3}{2}$

= 20( –$\frac{7}{2}$ ) + 390 ( $\frac{3}{2}$ )

= -70 + 585 = 515

Correct answer : Option (b)

Practice Questions :

1. In a geometric series of positive terms the difference between the fifth and fourth terms is 576, and the difference between the second and first terms is 9 what is the sum of the first five terms of this series ?

Options :

(A) 1061

(B) 1023

(C) 1024

(D) 768

(E) None of these

Answer :

Given

ar⁵ – ar⁴ = 576

ar⁴ ( r – 1 ) = 576 ——————– 1

ar² – a = 9

a ( r – 1) = 9 ——————— 2

Divide 1 / 2

$\frac{ar^4 [ r – 1] }{ar [ r -1]}$ = $\frac{576}{9}$

r³ = 64

r = 4

substitude r = 4 in 2nd equation

a (3) = 9

3a = 9

a = 3

S_n = $\frac{ a (r^n – 1)}{ r -1}$

S₅ = $\frac{ 3 ( 4^5 – 1)}{3}$

S₅ = $\frac{ 3 [ 1024 – 1]}{3}$

S₅ = 1023

Correct answer : Option (B)

2. For what values of x, the numbers $\frac{-2}{7}$, x, $\frac{-7}{2}$ are in G.P ?

Solution :

a₁ – $\frac{-2}{7}$, a₂ – x, a₃ – $\frac{-7}{2}$

we know that

$\frac{a_2}{a_1}$ = $\frac{a_3}{a_2}$

$\frac{x}{-2/7}$ = $\frac{-7/2}{x}$

x² = $\frac{-7}{2}$ x $\frac{-2}{7}$

x² = 1

x = $\pm\sqrt{1}$

x = $\pm$1

3. Which term of the following sequences 2, 2$\sqrt{2}$, 4……………………….. is 128 ?

Solution :

a = 2, r = $\frac{a_2}{a_1}$ = $\frac{2\sqrt{2}}{2}$ = $\sqrt{2}$

a_n = 128

(2)($\sqrt{2}$)^{n – 1} = 128

2. 2^{n – 1/2} = 2⁷

2^{1 + n/2} = 2⁷

If bases are equal then exponents are equal

$\frac{1 + n}{2}$ = 7

1 + n = 14

n = 14 – 1 = 13

128 is 13^th term

4. The 4^th term of a G.P is square of its second term and the first term is -3. Determine its 7^th term.

Solution :

a₄ = (a₂)²

(ar³) = (ar)²

(-3).r³ = (-3)².r²

$\frac{r^3}{r^2}$ = $\frac{(-3)^2}{-3}$

r = -3

a = -3

a₇ = ar⁶

= (-3)(-3)⁶

= (-3)⁷

= -2187

5. Which term of the following sequences $\sqrt{3}$, 3, 3$\sqrt{3}$,……………….. is 729 ?

Solution :

a = $\sqrt{3}$, r = $\frac{a_2}{a_1}$ = $\frac{3}{\sqrt{3}}$ = $\frac{(\sqrt{3})^2}{\sqrt{3}}$

a_n = 729

a_n = a.r^n-1

($\sqrt{3})^n$= 729 = 3⁶

(3)^n/2 = 3⁶

$\frac{n}{2}$ = 6

n = 12

729 is 12^th term

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