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