If the sum and product of four positive consecutive terms of a G.… — Solution | JEE Mathematics

MCQMediumJEE 2023Geometric Progression (GP)

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:

A
$7$
B
$\frac{9}{2}$
C
$3$
D
$14$

Answer

Correct answer:C

Step-by-step solution

Standard Method

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

Find: The sum of all possible common ratios $r$ .

Let the four consecutive terms be

a, ar, ar^2, ar^3

with $a>0$ and $r>0$ .

Using the product,

a \cdot ar \cdot ar^2 \cdot ar^3 = a^4 r^6 = 1296

So,

a^2 r^3 = 36

Using the transformed variable

From

a^2 r^3 = 36

we get

a = \frac{6}{r^{3/2}}

Convert the sum into a symmetric form

Now use the sum condition:

a + ar + ar^2 + ar^3 = 126

Substituting $a = \frac{6}{r^{3/2}}$ ,

\frac{6}{r^{3/2}} + \frac{6}{r^{1/2}} + 6r^{1/2} + 6r^{3/2} = 126

Dividing by $6$ ,

r^{-3/2} + r^{-1/2} + r^{1/2} + r^{3/2} = 21

Group the terms as

\left(r^{1/2}+r^{-1/2}\right) + \left(r^{3/2}+r^{-3/2}\right) = 21

Let

A = r^{1/2}+r^{-1/2}

Then

r^{3/2}+r^{-3/2} = A^3 - 3A

Hence,

A + (A^3-3A) = 21

A^3 - 2A = 21

From the working in the solution,

A = 3

So,

\sqrt{r} + \frac{1}{\sqrt{r}} = 3

Multiplying by $\sqrt{r}$ ,

r + 1 = 3\sqrt{r}

Squaring,

r^2 + 2r + 1 = 9r

r^2 - 7r + 1 = 0

Therefore, the two possible common ratios are the roots of this quadratic, and their sum is $7$ .

The solution marks option C, but the derived value is $7$ , which matches option A. Therefore, the defensible correct option is A.

Common mistakes

Taking the four terms as $a, a+r, a+2r, a+3r$ treats them as an A.P., not a G.P. This gives the wrong equations. Use $a, ar, ar^2, ar^3$ instead.
Using the product incorrectly as $a^4r^3$ is wrong because the powers of $r$ add to $0+1+2+3=6$ . The correct product is $a^4r^6$ .
After getting $\sqrt{r}+\frac{1}{\sqrt{r}}=3$ , forgetting to multiply by $\sqrt{r}$ before squaring can lead to algebra errors. First write $r+1=3\sqrt{r}$ , then square carefully.

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