JEE Main 2023 (31 January Shift 1)

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Animated Solution for Mathematics - Sequence and Series: 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

Q: Why do we reject the other roots of the cubic equation $x^3 - 2x - 21 = 0$?

Great question. When we factorized the cubic, we got $(x-3)(x^2 + 3x + 7) = 0$. The quadratic part $x^2 + 3x + 7$ has a discriminant $D = 3^2 - 4(1)(7) = 9 - 28 = -19$. Since the discriminant is negative, the roots are complex. In the context of this problem, we are dealing with a G.P. of positive terms, which implies $r$ must be a real number. Therefore, we only accept the real root $x=3$.

Select Answer:

Visualized Solution

Defining the G.P. Terms

Let the four consecutive terms of the G.P. be $a, a r, a r^{2}, a r^{3}$ .
Given that the terms are positive, we have $a > 0$ and $r > 0$ .

Using the Product Condition

Product of terms: $a \times a r \times a r^{2} \times a r^{3} = 1296$
$⟹ a^{4} r^{6} = 1296$

Simplifying the Product Equation

$(a^{2} r^{3})^{2} = 3 6^{2}$
$⟹ a^{2} r^{3} = 36$
$⟹ a = \frac{6}{r ^{3/2}}$

Using the Sum Condition

Sum of terms: $a + a r + a r^{2} + a r^{3} = 126$
$⟹ a (1 + r + r^{2} + r^{3}) = 126$

Substituting $a$ into the Sum

Substitute $a = \frac{6}{r ^{3/2}}$ :
$\frac{6}{r ^{3/2}} (1 + r + r^{2} + r^{3}) = 126$

Simplifying the Sum Expression

Divide by 6: $\frac{1 + r + r ^{2} + r ^{3}}{r ^{3/2}} = 21$
Separate the terms: $r^{- 3/2} + r^{- 1/2} + r^{1/2} + r^{3/2} = 21$

Symmetric Rearrangement

Group symmetric terms:
$(r^{3/2} + r^{- 3/2}) + (r^{1/2} + r^{- 1/2}) = 21$

Algebraic Substitution

Let $x = r^{1/2} + r^{- 1/2}$
Then $x^{3} = (r^{1/2} + r^{- 1/2})^{3} = r^{3/2} + r^{- 3/2} + 3 (r^{1/2} + r^{- 1/2})$
$⟹ r^{3/2} + r^{- 3/2} = x^{3} - 3 x$

Forming the Cubic Equation

Substitute back into the sum equation:
$(x^{3} - 3 x) + x = 21$
$⟹ x^{3} - 2 x - 21 = 0$

Solving for $x$

By inspection, $x = 3$ is a root since $3^{3} - 2 (3) - 21 = 27 - 6 - 21 = 0$ .
Factorizing: $(x - 3) (x^{2} + 3 x + 7) = 0$
Since $x^{2} + 3 x + 7 = 0$ has no real roots, $x = 3$ is the only real solution.

Finding the Common Ratio $r$

We have $r^{1/2} + r^{- 1/2} = 3$
Square both sides: $(r + \frac{1}{r})^{2} = 3^{2}$
$⟹ r + \frac{1}{r} + 2 = 9$
$⟹ r + \frac{1}{r} = 7$
$⟹ r^{2} - 7 r + 1 = 0$

Sum of Common Ratios

The possible values of the common ratio $r$ are the roots of $r^{2} - 7 r + 1 = 0$ .
Sum of roots $= - \frac{b}{a} = - \frac{- 7}{1} = 7$ .
The sum of all such common ratios is 7.

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The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

Geometric Progressions (G.P.) are the heartbeat of growth and decay in our universe. When you see a problem involving four consecutive terms of a G.P., denote them as

a, a r, a r^{2}, a r^{3}

We are given that these terms are positive, which implies

a > 0

and

r > 0

. This constraint serves as our anchor throughout the algebraic manipulation.

The Power of the Product

We are given that the product of these four terms is

1296

. Therefore:

a \times a r \times a r^{2} \times a r^{3} = 1296

Simplifying the expression, we obtain:

a^{4} r^{6} = 1296

Since

1296 = 3 6^{2}

, we can write

(a^{2} r^{3})^{2} = 3 6^{2}

. This simplifies to the fundamental relationship:

a^{2} r^{3} = 36 ⟹ a = \frac{6}{r ^{3/2}}

The Art of Substitution

Next, we consider the sum of the four terms, which is given as

126

a (1 + r + r^{2} + r^{3}) = 126

Substituting our expression for

a

into this equation yields:

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

Dividing both sides by

6

and distributing the denominator

r^{3/2}

, we get:

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

The Cubic Transformation

To solve this, we introduce the substitution

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

. Cubing this expression gives:

x^{3} = (r^{1/2} + r^{- 1/2})^{3} = r^{3/2} + r^{- 3/2} + 3 (r^{1/2} + r^{- 1/2}) = r^{3/2} + r^{- 3/2} + 3 x

Rearranging, we find

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

. Substituting this back into our sum equation:

(x^{3} - 3 x) + x = 21 ⟹ x^{3} - 2 x - 21 = 0

By testing integer roots, we find that

x = 3

satisfies the equation (

27 - 6 - 21 = 0

The Final Victory

With

x = 3

, we have

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

. Squaring both sides results in:

r + \frac{1}{r} + 2 = 9 ⟹ r + \frac{1}{r} = 7

This simplifies to the quadratic equation:

r^{2} - 7 r + 1 = 0

The roots of this quadratic represent the possible values of the common ratio

r

. By Vieta's formulas, the sum of these roots is:

Sum = - (\frac{- 7}{1}) = 7

Common Traps & FAQs

Wait, why did we assume the terms are

a, a r, a r^{2}, a r^{3}

? Could we have used

a / r^{3}, a / r, a r, a r^{3}

That is a brilliant observation! Yes, you absolutely could have used

a / r^{3}, a / r, a r, a r^{3}

(where the common ratio is

r^{2}

). This often simplifies the product calculation because the

r

terms cancel out. However, the standard

a, a r, a r^{2}, a r^{3}

approach works perfectly fine as long as you are careful with the algebra. Both paths lead to the same destination—the beauty of math is that there are many roads to the truth!

Why do we reject the other roots of the cubic equation

x^{3} - 2 x - 21 = 0

Great question. When we factorized the cubic, we got

(x - 3) (x^{2} + 3 x + 7) = 0

. The quadratic part

x^{2} + 3 x + 7

has a discriminant

D = 3^{2} - 4 (1) (7) = 9 - 28 = - 19

. Since the discriminant is negative, the roots are complex. In the context of this problem, we are dealing with a G.P. of positive terms, which implies

r

must be a real number. Therefore, we only accept the real root

x = 3

Why can't we just solve for

r

directly from the sum equation without the substitution?

You could try, but you would end up with a quartic equation (degree 4) in terms of

r

, which is much harder to solve directly. The substitution

x = r + \frac{1}{r}

is a classic JEE technique to reduce the degree of the equation. It transforms a complex-looking expression into a manageable cubic equation. Always look for symmetry!

Animated Solution for Mathematics - Sequence and Series: 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

Select Answer:

Visualized Solution

Defining the G.P. Terms

Using the Product Condition

Simplifying the Product Equation

Using the Sum Condition

Substituting $a$ into the Sum

Simplifying the Sum Expression

Symmetric Rearrangement

Algebraic Substitution

Forming the Cubic Equation

Solving for $x$

Finding the Common Ratio $r$

Sum of Common Ratios

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Power of the Product

The Art of Substitution

The Cubic Transformation

The Final Victory

Similar Questions

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

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In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals

Let $a_{1}, a_{2}, a_{3}, \dots$ be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $a_{1} a_{9} + a_{2} a_{4} a_{9} + a_{5} + a_{7}$ is equal to

Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is:

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The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then $2 α$ is

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 :

Animated Solution for Mathematics - Sequence and Series: 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

Select Answer:

Visualized Solution

Defining the G.P. Terms

Using the Product Condition

Simplifying the Product Equation

Using the Sum Condition

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a into the Sum

Simplifying the Sum Expression

Symmetric Rearrangement

Algebraic Substitution

Forming the Cubic Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Finding the Common Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Sum of Common Ratios

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Power of the Product

The Art of Substitution

The Cubic Transformation

The Final Victory

Similar Questions

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals

Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is

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 :

Substituting $a$ into the Sum

Solving for $x$

Finding the Common Ratio $r$

Sum of infinite number of terms of GP is $20$ and sum of their square is $100$ . The common ratio of GP is