JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Sequences and Series

Suppose that the number of terms in an A.P. is 2 k ,

k \in N

. If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to

(A)5✓ Correct
(B)8
(C)6
(D)4

Answer: (A)

Solution

Let the A.P. have

2 k

terms with first term

a_{1}

and common difference

d

.

The odd-positioned terms

a_{1}, a_{3}, \dots, a_{2 k - 1}

form an A.P. of

k

terms with common difference

2 d

sum = k [a_{1} + (k - 1) d] = 40.

The even-positioned terms

a_{2}, a_{4}, \dots, a_{2 k}

form an A.P. of

k

terms, first term

a_{1} + d

, common difference

2 d

sum = k [a_{1} + k d] = 55.

Subtracting:

k \cdot d = 55 - 40 = 15

, so

k d = 15

.

Also

a_{2 k} - a_{1} = (2 k - 1) d = 27

. Since

d = \frac{15}{k}

(2 k - 1) \cdot \frac{15}{k} = 27 \Rightarrow 30 k - 15 = 27 k \Rightarrow 3 k = 15 \Rightarrow k = 5.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Sequences and Series

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