Sequence and series JEE Main 2025 with solution

Question :

The number of terms of an AP is even. The sum of all the odd terms is 24 the sum of all the even term is 30 and the last term exceeds the first by $\frac{21}{2}$ then the number of terms which are integers in the AP

Options :

(A) 4

(B) 10

(C) 6

(D) 8

Sequence and series JEE Main 2025 with solution

Answer :

a₁, a₂, a₃, a₄ …………………… a_n-1, a_n

Sum of odd terms $\rightarrow$ a₁ + a₃ + a₅ + …………………….+ a_n-1 = 24 ————–1

Number of odd terms $\rightarrow$ $\frac{n}{2}$

Sum of even terms $\rightarrow$ a₂ + a₄ + a₆ + ……………………+ a_n = 30 —————–2

Number of even terms $\rightarrow$ $\frac{n}{2}$

Substract equation 2 – 1

( a₂ – a₁ ) + ( a₄ – a₃ ) + ( a₆ – a₅ ) + …………. = 6

d + d + d + d +…………………..+ d = 6

$\frac{n}{2}$(d) = 6

nd = 12

a_n = a + $\frac{21}{2}$ [ Given in question ]

a + ( n -1)d = a + $\frac{21}{2}$

nd – d = $\frac{21}{2}$

12 – d = $\frac{21}{2}$

-d = $\frac{21}{2}$ – 12

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

substitute d = $\frac{3}{2}$ in nd = 12

n x $\frac{3}{2}$ = 12

n = 8

a₁ + a₃ + a₅ + …………. = 24

d = 2d , $\frac{n}{2}$ terms

Using S_n formula

S_n = $\frac{n}{2}$[ 2a + ( n – 1)d ]

$\frac{n}{4}$ [ 2a + ( $\frac{n}{2}$ – 1 ) 2d ] = 24

n = 8 and d = $\frac{3}{2}$

2 [ 2a + 3(3) ] = 24

2a + 9 = 12

2a = 3

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

Total number of terms – $\frac{3}{2}$ , 3, $\frac{9}{2}$, 6 , $\frac{15}{2}$, 9 , $\frac{21}{2}$, 12

Number of integers = 4

Practice Questions :

1. The sum of first terms of a G.P is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P

Solution :

G.P $\rightarrow$ a, ar, ar², ar³, ……………………

Given

Sum of first terms is 16 $\rightarrow$ a + ar + ar² = 16

a [ 1 + r + r² ] = 16 ——————–1

Sum of next three terms is 128 $\rightarrow$ ar³ + ar⁴ + ar⁵ = 128

ar³ [ 1 + r + r² ] = 128 ————————–2

divide $\frac{2}{1}$ eq

$\frac{ar^3[ 1 + r + r^2 ]}{a[ 1 + r + r^2]}$ = $\frac{128}{16}$

r³ = 8

r = 2

Sub r in eq 1 $\rightarrow$ a [ 1 + r + r² ] = 16

a [ 1 + 2 + 4 ] = 16

a = $\frac{16}{7}$

S_n = $\frac{a(r^2 – 1)}{r – 1}$ = $\frac{16(2^n – 1)}{7(2 – 1)}$ = $\frac{16(2^n – 1)}{7}$

2. Given a G.P with a = 729 and 7^th term is 64, determine S₇

Solution :

a = 729, a₇ = 64

a_n = ar^{n – 1}

ar⁶ = 64

r⁶ = $\frac{64}{729}$

r⁶ = $\frac{2^6}{3^6}$

r = $6\sqrt{(\frac{2}{6})^6}$ = $\pm\frac{2}{3}$

If r = $\frac{2}{3}$

S₇ = $\frac{729[ 1 – \frac{2}{3}]^7}{ 1 – \frac{2}{3}}$ = $3^6\frac{3^7 – 2^7}{3^7} x \frac{3}{3 – 2}$ = 3⁷ – 2⁷ = 2187 – 128 = 2059

If r = $\frac{-2}{3}$

S₇ = $729\frac{ (1 – \frac{-2}{3}]^7}{1 – \frac{(-2}{3})}$ = $\frac{2187 + 128}{5}$ = 463

S₇ = 463

3. Find a G.P for which sum of the first two terms is -4 and the fifth term is 4 times the third term.

Solution :

Let the G.P be a, ar, ar², ar³,………………….

Given sum of first two terms is -4 $\rightarrow$ a + ar = -4

a( 1 + r) = -4

Given a₅ = 4a₃

ar⁴ = 4ar²

$\frac{ar^4}{ar^2}$ = 4

r² = 4

r = $\pm\sqrt{4}$

r = $\pm$2

If r = 2 eq 1

a[ 1 + 2] = – 4

a = $\frac{-4}{3}$

G.P is a, ar, ar², ar³, ……………

$\frac{-4}{3}$, $\frac{-4}{3}$(2), $\frac{-4}{3}$(2)², $\frac{-4}{3}$(2)³, ………………

$\frac{-4}{3}$, $\frac{-8}{3}$, $\frac{-16}{3}$, $\frac{-32}{3}$,………………………..

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