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JEE Mathematics 2025 Question with Solution

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 2121 and the sum of its eighth, tenth and twelfth terms is 1530915309, then the sum of its first nine terms is:

  • A

    760760

  • B

    755755

  • C

    750750

  • D

    757757

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The sum of the second, fourth and sixth terms of a G.P. is 2121, and the sum of the eighth, tenth and twelfth terms is 1530915309.

Find: The sum of the first nine terms.

Let the first term be aa and the common ratio be rr. Then the G.P. is a,ar,ar2,ar3,…a, ar, ar^2, ar^3, \ldots

The second, fourth and sixth terms are ar,ar3,ar5ar, ar^3, ar^5. So,

ar+ar3+ar5=21ar + ar^3 + ar^5 = 21

The eighth, tenth and twelfth terms are ar7,ar9,ar11ar^7, ar^9, ar^{11}. So,

ar7+ar9+ar11=15309ar^7 + ar^9 + ar^{11} = 15309

Divide the second equation by the first:

ar7+ar9+ar11ar+ar3+ar5=1530921\frac{ar^7 + ar^9 + ar^{11}}{ar + ar^3 + ar^5} = \frac{15309}{21}

This gives

r6⋅1+r2+r41+r2+r4=729r^6 \cdot \frac{1 + r^2 + r^4}{1 + r^2 + r^4} = 729

Hence,

r6=729=36r^6 = 729 = 3^6

Therefore,

r=3r = 3

since the terms are positive.

Now substitute r=3r = 3 into

ar+ar3+ar5=21ar + ar^3 + ar^5 = 21

We get

a(3)+a(33)+a(35)=21a(3) + a(3^3) + a(3^5) = 21 a(3+27+243)=21a(3 + 27 + 243) = 21 a⋅273=21a \cdot 273 = 21 a=21273=113a = \frac{21}{273} = \frac{1}{13}

Now use the sum of first nn terms of a G.P.:

Sn=arn−1r−1S_n = a \frac{r^n - 1}{r - 1}

For the first nine terms,

S9=113⋅39−13−1S_9 = \frac{1}{13} \cdot \frac{3^9 - 1}{3 - 1} S9=113⋅19683−12S_9 = \frac{1}{13} \cdot \frac{19683 - 1}{2} S9=113⋅196822S_9 = \frac{1}{13} \cdot \frac{19682}{2} S9=113⋅9841S_9 = \frac{1}{13} \cdot 9841 S9=757S_9 = 757

Therefore, the sum of the first nine terms is 757757. The correct option is D.

Factored Equation Method

Given:

T2+T4+T6=21T_2 + T_4 + T_6 = 21

and

T8+T10+T12=15309T_8 + T_{10} + T_{12} = 15309

Find: S9S_9

Using Tn=arn−1T_n = ar^{n-1},

T2=ar,T4=ar3,T6=ar5T_2 = ar, \quad T_4 = ar^3, \quad T_6 = ar^5

So,

ar+ar3+ar5=21ar + ar^3 + ar^5 = 21

Factor out arar:

ar(1+r2+r4)=21ar(1 + r^2 + r^4) = 21

Similarly,

T8=ar7,T10=ar9,T12=ar11T_8 = ar^7, \quad T_{10} = ar^9, \quad T_{12} = ar^{11}

Thus,

ar7+ar9+ar11=15309ar^7 + ar^9 + ar^{11} = 15309

Factor out ar7ar^7:

ar7(1+r2+r4)=15309ar^7(1 + r^2 + r^4) = 15309

Now divide:

ar7(1+r2+r4)ar(1+r2+r4)=1530921\frac{ar^7(1 + r^2 + r^4)}{ar(1 + r^2 + r^4)} = \frac{15309}{21}

Hence,

r6=729r^6 = 729

So,

r=3r = 3

Substitute into

ar(1+r2+r4)=21ar(1 + r^2 + r^4) = 21 a⋅3(1+9+81)=21a \cdot 3(1 + 9 + 81) = 21 a⋅3⋅91=21a \cdot 3 \cdot 91 = 21 a=113a = \frac{1}{13}

Now,

S9=a1−r91−rS_9 = a\frac{1-r^9}{1-r}

Substitute a=113a = \frac{1}{13} and r=3r = 3:

S9=113⋅1−391−3S_9 = \frac{1}{13} \cdot \frac{1-3^9}{1-3} S9=113⋅1−19683−2S_9 = \frac{1}{13} \cdot \frac{1-19683}{-2} S9=113⋅9841=757S_9 = \frac{1}{13} \cdot 9841 = 757

Therefore, the correct option is D.

Common mistakes

  • Students may divide the two given sums incorrectly and miss that the common factor 1+r2+r41 + r^2 + r^4 cancels. This is wrong because both expressions have the same bracketed factor. First factor each sum properly, then divide to get r6r^6 directly.

  • Students may take r=±3r = \pm 3 after solving r6=729r^6 = 729. This is wrong because the question states the G.P. has positive terms, so the common ratio must be positive here. Use r=3r = 3.

  • Students may use the sum formula for a G.P. with sign errors, such as mixing up rn−1r−1\frac{r^n-1}{r-1} and 1−rn1−r\frac{1-r^n}{1-r}. Both are equivalent, but only if used consistently. Choose one correct form and substitute carefully.

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