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JEE Mathematics 2024 Question with Solution

Let 2nd2^{\text{nd}}, 8th8^{\text{th}}, and 44th44^{\text{th}} terms of a non-constant A.P. be respectively the 1st1^{\text{st}}, 2nd2^{\text{nd}}, and 3rd3^{\text{rd}} terms of a G.P. If the first term of the A.P. is 11, then the sum of the first 2020 terms is equal to:

  • A

    980980

  • B

    960960

  • C

    990990

  • D

    970970

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The first term of the A.P. is a=1a = 1 and its common difference is dd. The 2nd2^{\text{nd}}, 8th8^{\text{th}}, and 44th44^{\text{th}} terms of this A.P. are the 1st1^{\text{st}}, 2nd2^{\text{nd}}, and 3rd3^{\text{rd}} terms of a G.P.

Find: The sum of the first 2020 terms of the A.P.

The corresponding terms of the A.P. are

a2=a+d=1+da_2 = a + d = 1 + d a8=a+7d=1+7da_8 = a + 7d = 1 + 7d a44=a+43d=1+43da_{44} = a + 43d = 1 + 43d

Since these three terms are in G.P., the square of the middle term equals the product of the other two:

(1+7d)2=(1+d)(1+43d)(1 + 7d)^2 = (1 + d)(1 + 43d)

Expanding both sides,

1+14d+49d2=1+44d+43d21 + 14d + 49d^2 = 1 + 44d + 43d^2

So,

49d2+14d=43d2+44d49d^2 + 14d = 43d^2 + 44d 6d2−30d=06d^2 - 30d = 0 6d(d−5)=06d(d - 5) = 0

Thus,

d=0ord=5d = 0 \quad \text{or} \quad d = 5

Since the A.P. is non-constant, d≠0d \ne 0. Therefore,

d=5d = 5

Now use the sum formula for an A.P.:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}\left[2a + (n-1)d\right]

For n=20n = 20, a=1a = 1, and d=5d = 5,

S20=202[2⋅1+(20−1)⋅5]S_{20} = \frac{20}{2}\left[2 \cdot 1 + (20 - 1) \cdot 5\right] S20=10(2+95)=10⋅97=970S_{20} = 10(2 + 95) = 10 \cdot 97 = 970

Therefore, the sum of the first 2020 terms is 970970. The correct option is D.

Use GP condition directly

Given: a=1a = 1 for the A.P., with common difference dd.

Find: The sum of the first 2020 terms.

Write the required A.P. terms directly as

1+d,1+7d,1+43d1+d, \quad 1+7d, \quad 1+43d

Because they are in G.P., use

(middle term)2=(first term)(third term)(\text{middle term})^2 = (\text{first term})(\text{third term})

So,

(1+7d)2=(1+d)(1+43d)(1+7d)^2 = (1+d)(1+43d)

This immediately gives

1+14d+49d2=1+44d+43d21 + 14d + 49d^2 = 1 + 44d + 43d^2 6d2−30d=06d^2 - 30d = 0 d(d−5)=0d(d-5)=0

Since the A.P. is non-constant, take d=5d=5.

Now compute

S20=202[2+19⋅5]=10(97)=970S_{20} = \frac{20}{2}[2 + 19 \cdot 5] = 10(97) = 970

Therefore, the correct option is D.

Common mistakes

  • Using the A.P. condition instead of the G.P. condition for a2,a8,a44a_2, a_8, a_{44} is incorrect. These terms are not in A.P.; they are given to be consecutive terms of a G.P. Use a82=a2a44a_8^2 = a_2 a_{44} instead.

  • Taking d=0d = 0 from 6d(d−5)=06d(d-5)=0 ignores the phrase non-constant A.P. A constant A.P. has common difference 00, so this value must be rejected.

  • Writing the 8th8^{\text{th}} or 44th44^{\text{th}} term incorrectly as 1+8d1+8d or 1+44d1+44d is a common indexing mistake. In an A.P., the nthn^{\text{th}} term is a+(n−1)da+(n-1)d, so these terms are 1+7d1+7d and 1+43d1+43d.

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