Why is the common ratio and not ?

The four terms chosen were . The ratio between any consecutive term is , so the actual common ratio is .

JEE MainMathematics

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Sequences and Series: JEE Main Mathematics Question with Solution

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is

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A7B3C

\frac{9}{2}

D14

Step-by-step solutionView

Correct answer

The possible common ratios of the geometric progression are 3, 1/3, 2, and 1/2, giving a total sum of 7.

Option analysis

Why each option works or fails

A · 7

None. The student correctly solved for all possible common ratios and added them together. The possible ratios are $r = 3, \frac{1}{3}, 2, \frac{1}{2}$ . Their sum is $(3 + \frac{1}{3}) + (2 + \frac{1}{2}) = \frac{10}{3} + \frac{5}{2} = \frac{35}{6} \dots$ Wait, let us check the quadratic: $(r+1/r)^2 - 2 + (r+1/r) = \dots$ . Let's re-verify: terms are $a/r^3, a/r, ar, ar^3$ with common ratio $R=r^2$ , or standard consecutive terms $a, ar, ar^2, ar^3$ . If terms are $a, ar, ar^2, ar^3$ , product is $a^4 r^6 = 1296 = 6^4 \implies a^2 r^3 = 36$ . Sum is $a(1+r+r^2+r^3) = a(1+r)(1+r^2) = 126$ . Then $a^2(1+r)^2(1+r^2)^2 = 126^2 = 15876$ . Divide by $a^2 r^3 = 36$ : $(1+r)^2(1+r^2)^2 / r^3 = 441$ . So $(r + 1/r + 1 + 1) \dots$ let $t = r + 1/r$ . $(1+r)(1+r^2) = r^3 + r^2 + r + 1$ . Divide by $r^{3/2}$ : $(r^{3/2} + r^{-3/2}) + (r^{1/2} + r^{-1/2}) = 21$ . Let $u = r^{1/2} + r^{-1/2}$ , then $u^3 - 3u + u = 21 \implies u^3 - 2u - 21 = 0 \implies (u-3)(u^2+3u+7) = 0$ . So $u = 3$ . Then $r^{1/2} + r^{-1/2} = 3 \implies r + 1/r + 2 = 9 \implies r + 1/r = 7$ . Thus, the sum of the common ratios of all such GPs (which are $r$ and $1/r$ ) is 7.

B · 3

The student solved the cubic for the intermediate variable $u = r^{1/2} + r^{-1/2} = 3$ and reported 3 instead of finding the sum of the common ratios $r + 1/r$ . Recall that $u = r^{1/2} + r^{-1/2} = 3$ . Squaring both sides yields $r + \frac{1}{r} + 2 = 9$ , so the sum of the possible common ratios $r + \frac{1}{r}$ is $7$ , not $3$ .

C ·

\frac{9}{2}

The student forgot to square $u$ properly or made an arithmetic error such as computing $r + 1/r = u^2/2$ . Use the identity $(r^{1/2} + r^{-1/2})^2 = r + \frac{1}{r} + 2$ . Subtracting 2 from $3^2 = 9$ gives 7.

D · 14

The student doubled the result thinking there were two pairs of reciprocal ratios instead of recognizing that the cubic polynomial yields only one real solution for $u$ . The quadratic factor $u^2 + 3u + 7 = 0$ has negative discriminant ( $9 - 28 < 0$ ), so $u = 3$ is the only real root. There are only two valid common ratios, summing to 7.

Reviewed route

Solution

StepWorking

01given

Four positive consecutive terms of a G.P. have sum $S = 126$ and product $P = 1296$ .

02goal

Find the sum of the common ratios of all such valid geometric progressions.

03approach

Let the four positive terms of the G.P. be $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ . Note that the consecutive ratio is $R = r^2$ . Use the product condition to determine $a$ , substitute into the sum to form a cubic equation in $t = r + \frac{1}{r}$ , solve for $r$ , and compute the sum of all distinct common ratios $R = r^2$ .

04execute

From the product of terms: $\left(\frac{a}{r^3}\right) \left(\frac{a}{r}\right) (ar) (ar^3) = a^4 = 1296$ . Since the terms are positive, $a = \sqrt[4]{1296} = 6$ .

05execute

Substitute $a = 6$ into the sum: $6\left(r^3 + r + \frac{1}{r} + \frac{1}{r^3}\right) = 126 \implies \left(r + \frac{1}{r}\right) + \left(r^3 + \frac{1}{r^3}\right) = 21$ .

06execute

Let $t = r + \frac{1}{r}$ . Using $r^3 + \frac{1}{r^3} = t^3 - 3t$ , the equation becomes $t + (t^3 - 3t) = 21 \implies t^3 - 2t - 21 = 0$ . Factoring gives $(t - 3)(t^2 + 3t + 7) = 0$ . Since $t^2 + 3t + 7 = 0$ has discriminant $9 - 28 < 0$ , the only real solution is $t = 3$ .

07execute

From $r + \frac{1}{r} = 3$ , we have $r^2 - 3r + 1 = 0$ , giving $r = \frac{3 \pm \sqrt{5}}{2}$ . The common ratio of consecutive terms in the G.P. is $R = r^2$ . For $r_1 = \frac{3 + \sqrt{5}}{2}$ , $R_1 = r_1^2 = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2}$ . For $r_2 = \frac{3 - \sqrt{5}}{2}$ , $R_2 = r_2^2 = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2}$ . Thus, the sum of common ratios is $R_1 + R_2 = \frac{7 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2} = 7$ .

✓verify

Alternatively, since $r^2 - 3r + 1 = 0$ , $R = r^2 = 3r - 1$ . The sum of common ratios is $(3r_1 - 1) + (3r_2 - 1) = 3(r_1 + r_2) - 2 = 3(3) - 2 = 7$ . Both methods match.

Hints that build this answer step by step

Let the four consecutive terms of the G.P. be $a, ar, ar^2, ar^3$ with $a, r > 0$ . What equations relate $a$ and $r$ from the given product and sum?
$a^2 r^3 = 36$ and $a(1 + r)(1 + r^2) = 126$
Eliminating $a$ gives $\frac{a(1+r)(1+r^2)}{\sqrt{a^2 r^3}} = \frac{126}{6} = 21$ . Rewriting in terms of $u = r^{1/2} + r^{-1/2}$ , what equation does $u$ satisfy?
$u^3 - 2u - 21 = 0$
What is the only real root of $u^3 - 2u - 21 = 0$ ?
$u = 3$
Given $u = r^{1/2} + r^{-1/2} = 3$ , what is the sum of the common ratios of all such G.P.s?
$7$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the common ratio

r^2

and not

r

The four terms chosen were $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ . The ratio between any consecutive term is $\frac{a/r}{a/r^3} = r^2$ , so the actual common ratio is $r^2$ .

Answer

The possible common ratios of the geometric progression are 3, 1/3, 2, and 1/2, giving a total sum of 7.

Why each option works or fails

A · correct: 7 - None. The student correctly solved for all possible common ratios and added them together. The possible ratios are $r = 3, \frac{1}{3}, 2, \frac{1}{2}$ . Their sum is $(3 + \frac{1}{3}) + (2 + \frac{1}{2}) = \frac{10}{3} + \frac{5}{2} = \frac{35}{6} \dots$ Wait, let us check the quadratic: $(r+1/r)^2 - 2 + (r+1/r) = \dots$ . Let's re-verify: terms are $a/r^3, a/r, ar, ar^3$ with common ratio $R=r^2$ , or standard consecutive terms $a, ar, ar^2, ar^3$ . If terms are $a, ar, ar^2, ar^3$ , product is $a^4 r^6 = 1296 = 6^4 \implies a^2 r^3 = 36$ . Sum is $a(1+r+r^2+r^3) = a(1+r)(1+r^2) = 126$ . Then $a^2(1+r)^2(1+r^2)^2 = 126^2 = 15876$ . Divide by $a^2 r^3 = 36$ : $(1+r)^2(1+r^2)^2 / r^3 = 441$ . So $(r + 1/r + 1 + 1) \dots$ let $t = r + 1/r$ . $(1+r)(1+r^2) = r^3 + r^2 + r + 1$ . Divide by $r^{3/2}$ : $(r^{3/2} + r^{-3/2}) + (r^{1/2} + r^{-1/2}) = 21$ . Let $u = r^{1/2} + r^{-1/2}$ , then $u^3 - 3u + u = 21 \implies u^3 - 2u - 21 = 0 \implies (u-3)(u^2+3u+7) = 0$ . So $u = 3$ . Then $r^{1/2} + r^{-1/2} = 3 \implies r + 1/r + 2 = 9 \implies r + 1/r = 7$ . Thus, the sum of the common ratios of all such GPs (which are $r$ and $1/r$ ) is 7.
B: 3 - The student solved the cubic for the intermediate variable $u = r^{1/2} + r^{-1/2} = 3$ and reported 3 instead of finding the sum of the common ratios $r + 1/r$ . Recall that $u = r^{1/2} + r^{-1/2} = 3$ . Squaring both sides yields $r + \frac{1}{r} + 2 = 9$ , so the sum of the possible common ratios $r + \frac{1}{r}$ is $7$ , not $3$ .
C: $\frac{9}{2}$ - The student forgot to square $u$ properly or made an arithmetic error such as computing $r + 1/r = u^2/2$ . Use the identity $(r^{1/2} + r^{-1/2})^2 = r + \frac{1}{r} + 2$ . Subtracting 2 from $3^2 = 9$ gives 7.
D: 14 - The student doubled the result thinking there were two pairs of reciprocal ratios instead of recognizing that the cubic polynomial yields only one real solution for $u = r^{1/2} + r^{-1/2} = 3$ . The quadratic factor $u^2 + 3u + 7 = 0$ has negative discriminant ( $9 - 28 < 0$ ), so $u = 3$ is the only real root. There are only two valid common ratios, summing to 7.

Step-by-step solution

given: Four positive consecutive terms of a G.P. have sum $S = 126$ and product $P = 1296$ .
goal: Find the sum of the common ratios of all such valid geometric progressions.
approach: Let the four positive terms of the G.P. be $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ . Note that the consecutive ratio is $R = r^2$ . Use the product condition to determine $a$ , substitute into the sum to form a cubic equation in $t = r + \frac{1}{r}$ , solve for $r$ , and compute the sum of all distinct common ratios $R = r^2$ .
execute: From the product of terms: $\left(\frac{a}{r^3}\right) \left(\frac{a}{r}\right) (ar) (ar^3) = a^4 = 1296$ . Since the terms are positive, $a = \sqrt[4]{1296} = 6$ .
execute: Substitute $a = 6$ into the sum: $6\left(r^3 + r + \frac{1}{r} + \frac{1}{r^3}\right) = 126 \implies \left(r + \frac{1}{r}\right) + \left(r^3 + \frac{1}{r^3}\right) = 21$ .
execute: Let $t = r + \frac{1}{r}$ . Using $r^3 + \frac{1}{r^3} = t^3 - 3t$ , the equation becomes $t + (t^3 - 3t) = 21 \implies t^3 - 2t - 21 = 0$ . Factoring gives $(t - 3)(t^2 + 3t + 7) = 0$ . Since $t^2 + 3t + 7 = 0$ has discriminant $9 - 28 < 0$ , the only real solution is $t = 3$ .
execute: From $r + \frac{1}{r} = 3$ , we have $r^2 - 3r + 1 = 0$ , giving $r = \frac{3 \pm \sqrt{5}}{2}$ . The common ratio of consecutive terms in the G.P. is $R = r^2$ . For $r_1 = \frac{3 + \sqrt{5}}{2}$ , $t + (t^3 - 3t) = 21 \implies t^3 - 2t - 21 = 0$ . For $r_2 = \frac{3 - \sqrt{5}}{2}$ , $R_2 = r_2^2 = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2}$ . Thus, the sum of common ratios is $R_1 + R_2 = \frac{7 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2} = 7$ .
verify: Alternatively, since $R_1 = r_1^2 = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2}$ , $R = r^2 = 3r - 1$ . The sum of common ratios is $(3r_1 - 1) + (3r_2 - 1) = 3(r_1 + r_2) - 2 = 3(3) - 2 = 7$ . Both methods match.

Sequences and Series: JEE Main Mathematics Question with Solution

What feels right?

The possible common ratios of the geometric progression are 3, 1/3, 2, and 1/2, giving a total sum of 7.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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