Maths Q7 – JEE-MAIN 2025

Home Maths Sequences and SeriesQuestion #414

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

Clue 1 of 3: Understand the structure of a G

P. Recall that the $n$ -th term of a G.P. is given by $a r^{n - 1}$ , where $a$ is the first term and $r$ is the common ratio.

🧠 Full Solution Path

Step 1: Set up equations from given information✦ Active

Let the first term of the G.P. be $a$ and the common ratio be $r$ . The terms are positive, so $a > 0$ and $r > 0$ . The given conditions can be written as:

a r + a r^{3} + a r^{5} = 21 (1)

a r^{7} + a r^{9} + a r^{11} = 15309 (2)

💡 Teacher's Secret Hint

Factor out common terms in each equation to simplify them.

Step 2: Solve for common ratio and first term○ Expand

Factor out $a r$ from (1) and $a r^{7}$ from (2):

a r (1 + r^{2} + r^{4}) = 21 (1^{'})

a r^{7} (1 + r^{2} + r^{4}) = 15309 (2^{'})

Divide equation (2') by equation (1'):

\frac{a r^{7} (1 + r^{2} + r^{4})}{a r (1 + r^{2} + r^{4})} = \frac{15309}{21}

r^{6} = 729

r = 3 (s i n c e r > 0)

Substitute $r = 3$ into equation (1'):

a (3) (1 + 3^{2} + 3^{4}) = 21

3 a (1 + 9 + 81) = 21

3 a (91) = 21

273 a = 21

a = \frac{21}{273} = \frac{1}{13}

💡 Teacher's Secret Hint

Remember that $r$ must be positive since all terms of the G.P. are positive.

Step 3: Calculate the sum of the first nine terms○ Expand

The sum of the first $n$ terms of a G.P. is given by $S_{n} = a \frac{r^{n} - 1}{r - 1}$ . For $n = 9$ , $a = \frac{1}{13}$ , and $r = 3$ :

S_{9} = \frac{1}{13} \frac{3^{9} - 1}{3 - 1}

S_{9} = \frac{1}{13} \frac{19683 - 1}{2}

S_{9} = \frac{1}{13} \frac{19682}{2}

S_{9} = \frac{1}{13} \times 9841

S_{9} = 757

💡 Teacher's Secret Hint

Be careful with the calculation of $3^{9}$ and the final division.

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Maths Question 7 – JEE-MAIN 2025

Clue 1 of 3: Understand the structure of a G

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