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MCQMediumJEE 2024Arithmetic Progression (AP)

JEE Mathematics 2024 Question with Solution

Let S(n)S(n) denote the sum of the first nn terms of an arithmetic progression. If S(10)=390S(10) = 390 and the ratio of the tenth and fifth terms is 15:715:7, then S(15)−S(5)S(15) - S(5) is equal to:

  • A

    800800

  • B

    890890

  • C

    790790

  • D

    690690

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: S10=390S_{10} = 390 and T10T5=157\frac{T_{10}}{T_5} = \frac{15}{7}.

Find: S15−S5S_{15} - S_5 for the arithmetic progression.

Use the sum formula of an AP:

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

For n=10n = 10,

102(2a+9d)=390\frac{10}{2}(2a + 9d) = 390 5(2a+9d)=3905(2a + 9d) = 390 2a+9d=78(1)2a + 9d = 78 \quad \text{(1)}

The nnth term of an AP is

Tn=a+(n−1)dT_n = a + (n-1)d

So,

T10=a+9d,T5=a+4dT_{10} = a + 9d, \qquad T_5 = a + 4d

Given

a+9da+4d=157\frac{a + 9d}{a + 4d} = \frac{15}{7}

Cross-multiplying,

7(a+9d)=15(a+4d)7(a + 9d) = 15(a + 4d) 7a+63d=15a+60d7a + 63d = 15a + 60d 8a=3d(2)8a = 3d \quad \text{(2)}

From equation (2)\text{(2)},

a=38da = \frac{3}{8}d

Substitute into equation (1)\text{(1)}:

2(38d)+9d=782\left(\frac{3}{8}d\right) + 9d = 78 34d+9d=78\frac{3}{4}d + 9d = 78 39d4=78\frac{39d}{4} = 78 39d=31239d = 312 d=8d = 8

Hence,

a=38×8=3a = \frac{3}{8} \times 8 = 3

Now compute the required sums:

S15=152(2a+14d)S_{15} = \frac{15}{2}(2a + 14d) S15=152(2×3+14×8)S_{15} = \frac{15}{2}(2 \times 3 + 14 \times 8) S15=152(6+112)S_{15} = \frac{15}{2}(6 + 112) S15=152×118=885S_{15} = \frac{15}{2} \times 118 = 885

Also,

S5=52(2a+4d)S_5 = \frac{5}{2}(2a + 4d) S5=52(2×3+4×8)S_5 = \frac{5}{2}(2 \times 3 + 4 \times 8) S5=52(6+32)=95S_5 = \frac{5}{2}(6 + 32) = 95

Therefore,

S15−S5=885−95=790S_{15} - S_5 = 885 - 95 = 790

The correct option is C.

Alternative Algebraic Method

Given: S10=390S_{10} = 390 and T10T5=157\frac{T_{10}}{T_5} = \frac{15}{7}.

Find: S15−S5S_{15} - S_5.

From

S10=102[2a+9d]=390S_{10} = \frac{10}{2}[2a + 9d] = 390

we get

2a+9d=782a + 9d = 78

The term ratio gives

a+9da+4d=157\frac{a+9d}{a+4d} = \frac{15}{7}

so

7(a+9d)=15(a+4d)7(a+9d) = 15(a+4d) 7a+63d=15a+60d7a + 63d = 15a + 60d 8a=3d  ⟹  d=8a38a = 3d \implies d = \frac{8a}{3}

Substitute d=8a3d = \frac{8a}{3} into

2a+9d=782a + 9d = 78

Then

2a+9(8a3)=782a + 9\left(\frac{8a}{3}\right) = 78 2a+24a=782a + 24a = 78 26a=7826a = 78 a=3a = 3

Hence,

d=8×33=8d = \frac{8 \times 3}{3} = 8

Now,

S15=152[2(3)+14(8)]=885S_{15} = \frac{15}{2}[2(3) + 14(8)] = 885

and

S5=52[2(3)+4(8)]=95S_5 = \frac{5}{2}[2(3) + 4(8)] = 95

Thus,

S15−S5=885−95=790S_{15} - S_5 = 885 - 95 = 790

Therefore, the value is 790790, so the correct option is C.

Common mistakes

  • Using the wrong AP sum formula, such as replacing 2a+(n−1)d2a + (n-1)d with a+nda + nd, gives an incorrect equation for S10S_{10}. Always use Sn=n2(2a+(n−1)d)S_n = \frac{n}{2}(2a + (n-1)d) exactly.

  • Writing the tenth and fifth terms incorrectly as a+10da + 10d and a+5da + 5d is wrong because the nnth term is a+(n−1)da + (n-1)d. So T10=a+9dT_{10} = a + 9d and T5=a+4dT_5 = a + 4d.

  • While using the ratio T10T5=157\frac{T_{10}}{T_5} = \frac{15}{7}, students often cross-multiply incorrectly or lose signs during rearrangement. Expand both sides carefully before simplifying to 8a=3d8a = 3d.

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