Logo
Jeeify
MCQMediumJEE 2026Arithmetic Progression (AP)

JEE Mathematics 2026 Question with Solution

If the sum of the first four terms of an A.P. is 66 and the sum of its first six terms is 44, then the sum of its first twelve terms is

  • A

    −22-22

  • B

    −20-20

  • C

    −26-26

  • D

    −24-24

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The sum of the first four terms of an A.P. is 66 and the sum of the first six terms is 44.

Find: The sum of the first twelve terms.

Let the first term be aa and the common difference be dd. Using the sum formula of an A.P.,

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

From S4=6S_4 = 6,

42[2a+3d]=6\frac{4}{2}[2a + 3d] = 6 2(2a+3d)=62(2a + 3d) = 6 2a+3d=32a + 3d = 3

From S6=4S_6 = 4,

62[2a+5d]=4\frac{6}{2}[2a + 5d] = 4 3(2a+5d)=43(2a + 5d) = 4 2a+5d=432a + 5d = \frac{4}{3}

Subtract the two equations:

(2a+5d)−(2a+3d)=43−3(2a + 5d) - (2a + 3d) = \frac{4}{3} - 3 2d=−532d = -\frac{5}{3} d=−56d = -\frac{5}{6}

Substitute in 2a+3d=32a + 3d = 3:

2a+3(−56)=32a + 3\left(-\frac{5}{6}\right) = 3 2a−52=32a - \frac{5}{2} = 3 2a=1122a = \frac{11}{2} a=114a = \frac{11}{4}

Now find S12S_{12}:

S12=122[2a+11d]S_{12} = \frac{12}{2}\left[2a + 11d\right] S12=6[2(114)+11(−56)]S_{12} = 6\left[2\left(\frac{11}{4}\right) + 11\left(-\frac{5}{6}\right)\right] S12=6(112−556)S_{12} = 6\left(\frac{11}{2} - \frac{55}{6}\right) S12=6(−113)=−22S_{12} = 6\left(-\frac{11}{3}\right) = -22

The working in the provided solution has an arithmetic slip in the final subtraction. Using the equations formed from the given sums, the correct value is −22-22. Therefore, the correct option is A.

Using linear equations in sums

Given: S4=6S_4 = 6 and S6=4S_6 = 4 for an A.P.

Find: S12S_{12}

Write the two sum expressions directly:

S4=2(2a+3d)=6S_4 = 2(2a+3d) = 6 S6=3(2a+5d)=4S_6 = 3(2a+5d) = 4

So,

2a+3d=3,2a+5d=432a+3d=3, \qquad 2a+5d=\frac{4}{3}

Subtracting gives

2d=43−3=−532d = \frac{4}{3}-3 = -\frac{5}{3} d=−56d = -\frac{5}{6}

Then

2a=3−3d=3+52=1122a = 3 - 3d = 3 + \frac{5}{2} = \frac{11}{2} a=114a = \frac{11}{4}

Now,

2a+11d=112−556=33−556=−226=−1132a + 11d = \frac{11}{2} - \frac{55}{6} = \frac{33-55}{6} = -\frac{22}{6} = -\frac{11}{3}

Hence,

S12=6(−113)=−22S_{12} = 6\left(-\frac{11}{3}\right) = -22

So the correct sum is −22-22.

Common mistakes

  • Using the formula for the nnth term instead of the sum formula is incorrect because the data given are values of S4S_4 and S6S_6. Start with Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a+(n-1)d], not an=a+(n−1)da_n = a+(n-1)d.

  • Forgetting to divide by the coefficient while forming equations leads to wrong linear equations. From S4=6S_4=6, you must get 2(2a+3d)=62(2a+3d)=6 and then 2a+3d=32a+3d=3.

  • A common error is the final fraction subtraction. Here 112−556=33−556=−226=−113\frac{11}{2} - \frac{55}{6} = \frac{33-55}{6} = -\frac{22}{6} = -\frac{11}{3}, not −23-\frac{2}{3}. Use a common denominator carefully before multiplying by 66.

Practice more Arithmetic Progression (AP) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Arithmetic Progression (AP)

Related questions