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JEE Mathematics 2025 Question with Solution

Let ana_n be the nn-th term of an A.P. If Sn=a1+a2+a3+⋯+an=700S_n = a_1 + a_2 + a_3 + \cdots + a_n = 700, a6=7a_6 = 7, and S7=7S_7 = 7, then ana_n is equal to:

  • A

    5656

  • B

    6565

  • C

    6464

  • D

    7070

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: a6=7a_6 = 7, S7=7S_7 = 7, and Sn=700S_n = 700 for an A.P.

Find: ana_n.

Use the A.P. formulas

ak=a1+(k−1)da_k = a_1 + (k-1)d

and

Sn=n2[2a1+(n−1)d].S_n = \frac{n}{2}\left[2a_1 + (n-1)d\right].

From a6=7a_6 = 7,

a1+5d=7(1)a_1 + 5d = 7 \qquad \text{(1)}

From S7=7S_7 = 7,

72[2a1+6d]=7\frac{7}{2}\left[2a_1 + 6d\right] = 7

so

2a1+6d=22a_1 + 6d = 2

and hence

a1+3d=1(2)a_1 + 3d = 1 \qquad \text{(2)}

Subtract (2) from (1):

(a1+5d)−(a1+3d)=7−1(a_1 + 5d) - (a_1 + 3d) = 7 - 1 2d=62d = 6 d=3d = 3

Then from (2),

a1+9=1a_1 + 9 = 1 a1=−8a_1 = -8

Now use Sn=700S_n = 700:

n2[2(−8)+(n−1)3]=700\frac{n}{2}\left[2(-8) + (n-1)3\right] = 700 n2(3n−19)=700\frac{n}{2}(3n - 19) = 700 n(3n−19)=1400n(3n - 19) = 1400 3n2−19n−1400=03n^2 - 19n - 1400 = 0

Solving the quadratic,

n=19±171616=19±1316n = \frac{19 \pm \sqrt{17161}}{6} = \frac{19 \pm 131}{6}

The feasible value is

n=25.n = 25.

Therefore,

an=a25=a1+(25−1)d=−8+24×3=64.a_n = a_{25} = a_1 + (25-1)d = -8 + 24 \times 3 = 64.

The correct option is C.

Using sum and term equations carefully

Given: a6=7a_6 = 7, S7=7S_7 = 7, and Sn=700S_n = 700.

Find: the value of ana_n.

The solution contains two approaches. In the first approach, the step from

S7=72(2a+6d)=7S_7 = \frac{7}{2}(2a + 6d) = 7

to a+3d=2a + 3d = 2 is inconsistent. Dividing correctly gives

2a1+6d=2⇒a1+3d=1.2a_1 + 6d = 2 \Rightarrow a_1 + 3d = 1.

The second approach uses the correct algebra, and that is the one followed.

Using the general term,

a6=a1+5d=7.a_6 = a_1 + 5d = 7.

Using the sum of first seven terms,

S7=72[2a1+6d]=7.S_7 = \frac{7}{2}[2a_1 + 6d] = 7.

Therefore,

2a1+6d=22a_1 + 6d = 2 a1+3d=1.a_1 + 3d = 1.

Now solve the pair:

a1+5d=7a_1 + 5d = 7 a1+3d=1a_1 + 3d = 1

Subtracting,

2d=6⇒d=3.2d = 6 \Rightarrow d = 3.

Then

a1=1−3d=1−9=−8.a_1 = 1 - 3d = 1 - 9 = -8.

Substitute in the sum formula:

700=n2[2(−8)+(n−1)3]700 = \frac{n}{2}[2(-8) + (n-1)3] 700=n2(−16+3n−3)700 = \frac{n}{2}(-16 + 3n - 3) 700=n2(3n−19)700 = \frac{n}{2}(3n - 19) 1400=n(3n−19)1400 = n(3n - 19) 3n2−19n−1400=0.3n^2 - 19n - 1400 = 0.

Factor or solve the quadratic to get the positive value

n=25.n = 25.

Hence,

an=a25=−8+24⋅3=64.a_n = a_{25} = -8 + 24 \cdot 3 = 64.

Therefore, the value of ana_n is 6464, so the correct option is C.

Common mistakes

  • Using S7=72(2a1+6d)=7S_7 = \frac{7}{2}(2a_1 + 6d) = 7 and simplifying to a1+3d=2a_1 + 3d = 2 is wrong. After dividing by 72\frac{7}{2}, we get 2a1+6d=22a_1 + 6d = 2, so the correct equation is a1+3d=1a_1 + 3d = 1. Always divide carefully before reducing.

  • Confusing ana_n with the first term a1a_1 leads to setting the wrong target. Here, first find a1a_1 and dd, then determine the specific value of nn from Sn=700S_n = 700, and only after that compute the required term ana_n.

  • Taking both roots of the quadratic for nn is incorrect. Since nn is the number of terms in a sum, it must be a positive integer. Discard the negative value and keep the feasible root only.

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