JEE MainMathematics

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Finding the Sum of Possible Common Ratios of a Geometric Progression

Q: Why is the common ratio r^2 and not r?

The four terms chosen were a/(r^3), a/r, ar, ar^3. The ratio between any consecutive term is (a/r)/(a/r^3) = r^2, so the actual common ratio is r^2.

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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A7B3C9/2D14

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

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 $a / r^{3}, a / r, a r, a r^{3}$ with common ratio $R = r^{2}$ , or standard consecutive terms $a, a r, a r^{2}, a r^{3}$ . If terms are $a, a r, a r^{2}, a r^{3}$ , product is $a^{4} r^{6} = 1296 = 6^{4} ⟹ a^{2} r^{3} = 36$ . 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 $u^{3} - 3 u + u = 21 ⟹ u^{3} - 2 u - 21 = 0 ⟹ (u - 3) (u^{2} + 3 u + 7) = 0$ . So $u = 3$ . Then $r^{1 / 2} + r^{- 1 / 2} = 3 ⟹ r + 1 / r + 2 = 9 ⟹ r + 1 / r = 7$ . 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 · 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 $u^{2} + 3 u + 7 = 0$ 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 $\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}$ . 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: $(\frac{a}{r^{3}}) (\frac{a}{r}) (a r) (a r^{3}) = a^{4} = 1296$ . Since the terms are positive, $a = \sqrt[4]{1296} = 6$ .

05Execute

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

06Execute

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

07Execute

From $r + \frac{1}{r} = 3$ , we have $r^{2} - 3 r + 1 = 0$ , 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 $r^{2} - 3 r + 1 = 0$ , $R = r^{2} = 3 r - 1$ . The sum of common ratios is $(3 r_{1} - 1) + (3 r_{2} - 1) = 3 (r_{1} + r_{2}) - 2 = 3 (3) - 2 = 7$ . Both methods match.

Hints that build this answer step by step

Let the four consecutive terms of the G.P. be $a, a r, a r^{2}, a r^{3}$ 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?
$u^{3} - 2 u - 21 = 0$
What is the only real root of $u^{3} - 2 u - 21 = 0$ ?
$u = 3$
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
Academic status: Answer verified against the official NTA key
Source: Previous-year question
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 $\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}$ . 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 - 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 $a / r^{3}, a / r, a r, a r^{3}$ with common ratio $R = r^{2}$ , or standard consecutive terms $a, a r, a r^{2}, a r^{3}$ . If terms are $a, a r, a r^{2}, a r^{3}$ , product is $a^{4} r^{6} = 1296 = 6^{4} ⟹ a^{2} r^{3} = 36$ . 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 $u^{3} - 3 u + u = 21 ⟹ u^{3} - 2 u - 21 = 0 ⟹ (u - 3) (u^{2} + 3 u + 7) = 0$ . So $u = 3$ . Then $r^{1 / 2} + r^{- 1 / 2} = 3 ⟹ r + 1 / r + 2 = 9 ⟹ r + 1 / r = 7$ . 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: 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 $1 / r$ . The quadratic factor $u^{2} + 3 u + 7 = 0$ 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 $u = r^{1 / 2} + r^{- 1 / 2} = 3$ 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 $\frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}$ . 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: $(\frac{a}{r^{3}}) (\frac{a}{r}) (a r) (a r^{3}) = a^{4} = 1296$ . Since the terms are positive, $a = \sqrt[4]{1296} = 6$ .
Execute: Substitute $a = 6$ into the sum: $6 (r^{3} + r + \frac{1}{r} + \frac{1}{r^{3}}) = 126 ⟹ (r + \frac{1}{r}) + (r^{3} + \frac{1}{r^{3}}) = 21$ .
Execute: Let $t = r + \frac{1}{r}$ . Using $r^{3} + \frac{1}{r^{3}} = t^{3} - 3 t$ , the equation becomes $t + (t^{3} - 3 t) = 21 ⟹ t^{3} - 2 t - 21 = 0$ . Factoring gives $(t - 3) (t^{2} + 3 t + 7) = 0$ . Since $t^{2} + 3 t + 7 = 0$ has discriminant $9 - 28 < 0$ , the only real solution is $t = 3$ .
Execute: From $r + \frac{1}{r} = 3$ , we have $r^{2} - 3 r + 1 = 0$ , 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 $r^{2} - 3 r + 1 = 0$ , $R = r^{2} = 3 r - 1$ . The sum of common ratios is $(3 r_{1} - 1) + (3 r_{2} - 1) = 3 (r_{1} + r_{2}) - 2 = 3 (3) - 2 = 7$ . Both methods match.

Finding the Sum of Possible Common Ratios of a Geometric Progression

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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