JEE Main 2020 (6 Sep Morning)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} -2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0$ . Then :

Q: What if $p$ was a complex number? Would the logic still hold?

Great question! The logic relies entirely on the fact that $a, b, c, d, p$ are real numbers. If $p$ were complex, the square of a number (like $(ap-b)^2$) could be negative, and the sum of squares could be zero without the individual terms being zero. The 'sum of squares equals zero implies each term is zero' rule is a special privilege of the real number system. Always check your domain constraints!

Q: I see the G.P. part, but why is it $a, b, c, d$ in G.P. and not $a, c, p$?

Look closely at the result of our equations: $\frac{b}{a} = p$, $\frac{c}{b} = p$, and $\frac{d}{c} = p$. This means the ratio between consecutive terms is constant ($p$). By definition, a sequence where the ratio of any term to its predecessor is constant is a Geometric Progression. Since the ratio is constant for the sequence $a, b, c, d$, they form the G.P.

Select Answer:

Progressive Hints

Hint 1

Look at the structure of the equation:

(a^{2} + b^{2} + c^{2}) p^{2} - 2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0

. It is a quadratic in

p

. When you see coefficients that look like sums of squares, your intuition should immediately scream 'perfect square completion'. Don't be intimidated by the variables; treat

p

as your variable and the rest as constants.

Hint 2

Recall the fundamental property of real numbers: if the sum of squares of real numbers is zero, i.e.,

x^{2} + y^{2} + z^{2} = 0

, then each term must individually be zero (

x = 0, y = 0, z = 0

). This is the 'trap-door' that unlocks the entire problem.

Hint 3

Start by expanding the expression:

a^{2} p^{2} + b^{2} p^{2} + c^{2} p^{2} - 2 ab p - 2 b c p - 2 c d p + b^{2} + c^{2} + d^{2} = 0

. Now, rearrange these terms to form three distinct perfect squares:

(a p - b)^{2} + (b p - c)^{2} + (c p - d)^{2} = 0

. From here, the path to the Geometric Progression is clear.

Visualized Solution

Analyzing the Given Equation

Given equation: $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0$
Our goal is to find the relationship between $a, b, c, d$ .

Expanding the Expression

Expand the terms by multiplying $p^{2}$ and $- 2 p$ :
$(a^{2} p^{2} + b^{2} p^{2} + c^{2} p^{2}) - (2 ab p + 2 b c p + 2 c d p) + (b^{2} + c^{2} + d^{2}) = 0$

Grouping for Perfect Squares

Rearrange and group the terms strategically:
$(a^{2} p^{2} - 2 ab p + b^{2}) + (b^{2} p^{2} - 2 b c p + c^{2}) + (c^{2} p^{2} - 2 c d p + d^{2}) = 0$

Rewriting as Sum of Squares

Using the algebraic identity $(x - y)^{2} = x^{2} - 2 x y + y^{2}$ :
$(a p - b)^{2} + (b p - c)^{2} + (c p - d)^{2} = 0$

Logic of Non-negative Terms

Since $a, b, c, d, p$ are real numbers, their squares are non-negative:
$(a p - b)^{2} \geq 0$ , $(b p - c)^{2} \geq 0$ , and $(c p - d)^{2} \geq 0$
The sum of non-negative terms is zero if and only if each term is exactly zero.

Equating Each Term to Zero

Set each individual squared term to zero:
$(a p - b) = 0$
$(b p - c) = 0$
$(c p - d) = 0$

Solving for the Common Ratio

Solve for $p$ in each equation:
$p = \frac{b}{a}$
$p = \frac{c}{b}$
$p = \frac{d}{c}$
Equating all of them gives: $\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$

Conclusion: Geometric Progression

Since the ratio of consecutive terms is constant:
$\frac{b}{a} = \frac{c}{b} = \frac{d}{c}$
This satisfies the definition of a Geometric Progression (G.P.).
Final Result: $a, b, c, d$ are in G.P.

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

Analyzing the Setup

We are given the quadratic equation:

(a^{2} + b^{2} + c^{2}) p^{2} - 2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0

At first glance, this expression appears to be a chaotic collection of variables. However, in the context of JEE Advanced, such structures often hide a deeper, more elegant algebraic identity.

The First Step

Expansion

We begin by expanding the expression to isolate the individual components. By distributing

p^{2}

and

- 2 p

across their respective brackets, we obtain:

(a^{2} p^{2} + b^{2} p^{2} + c^{2} p^{2}) - (2 ab p + 2 b c p + 2 c d p) + (b^{2} + c^{2} + d^{2}) = 0

Observe the terms carefully. We have

a^{2} p^{2}

b^{2}

, and a cross-term

- 2 ab p

. This structure strongly suggests the presence of the perfect square identity

(x - y)^{2} = x^{2} - 2 x y + y^{2}

The Elegant Grouping

To reveal the hidden structure, we rearrange the terms strategically by grouping the

a, b

terms, the

b, c

terms, and the

c, d

terms:

(a^{2} p^{2} - 2 ab p + b^{2}) + (b^{2} p^{2} - 2 b c p + c^{2}) + (c^{2} p^{2} - 2 c d p + d^{2}) = 0

The chaos now vanishes as each group forms a perfect square. The equation collapses into the following form:

(a p - b)^{2} + (b p - c)^{2} + (c p - d)^{2} = 0

The Logic of Zero

We now apply a fundamental property of real numbers: for any real number

x

x^{2} \geq 0

. If a sum of non-negative terms equals zero, each individual term must necessarily be zero.

If even one term were positive, the entire sum would exceed zero. Therefore, we must satisfy the following system of equations:

(a p - b) = 0, (b p - c) = 0, (c p - d) = 0

The Final Revelation

From these equations, we derive the ratios:

p = \frac{b}{a}, p = \frac{c}{b}, p = \frac{d}{c}

By equating these, we find:

\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p

This is the definition of a Geometric Progression (G.P.), where the ratio of consecutive terms is constant. We have successfully decoded the hidden relationship: $a, b, c, d$ are in G.P.

Common Traps & FAQs

Wait, why do we expand the whole thing? Can't we just use the discriminant

D \geq 0

condition for a quadratic equation?

That is a brilliant instinct! Yes, you absolutely could use the discriminant condition

D \leq 0

(since the quadratic must have real roots for

p

to exist, and the sum of squares implies the roots are real). However, expanding and grouping into perfect squares is often faster and more elegant because it reveals the specific relationship between the variables immediately, rather than just telling you that roots exist. Both methods are valid, but the 'sum of squares' method is the 'surgical' approach.

What if

p

was a complex number? Would the logic still hold?

Great question! The logic relies entirely on the fact that

a, b, c, d, p

are real numbers. If

p

were complex, the square of a number (like

(a p - b)^{2}

) could be negative, and the sum of squares could be zero without the individual terms being zero. The 'sum of squares equals zero implies each term is zero' rule is a special privilege of the real number system. Always check your domain constraints!

I see the G.P. part, but why is it

a, b, c, d

in G.P. and not

a, c, p

Look closely at the result of our equations:

\frac{b}{a} = p

\frac{c}{b} = p

, and

\frac{d}{c} = p

. This means the ratio between consecutive terms is constant (

p

). By definition, a sequence where the ratio of any term to its predecessor is constant is a Geometric Progression. Since the ratio is constant for the sequence

a, b, c, d

, they form the G.P.

Animated Solution for Mathematics - Sequence and Series: Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} -2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0$ . Then :

Select Answer:

Visualized Solution

Analyzing the Given Equation

Expanding the Expression

Grouping for Perfect Squares

Rewriting as Sum of Squares

Logic of Non-negative Terms

Equating Each Term to Zero

Solving for the Common Ratio

Conclusion: Geometric Progression

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

Analyzing the Setup

The First Step

The Elegant Grouping

The Logic of Zero

The Final Revelation

Similar Questions

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0$ then $a, b, c, d$

If three distinct numbers a,b,c are in G.P. and the equations $a x^{2} + 2 b x + c = 0$ and $d x^{2} + 2 e x + f = 0$ have a common root, then which one of the following statements is correct?

If a, b, c are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. such that $a < b < c$ and $a + b + c = \frac{3}{4}$ , then the value of a is :-

If $a_{1} (> 0), a_{2}, a_{3}, a_{4}, a_{5}$ are in a G.P., $a_{2} + a_{4} = 2 a_{3} + 1$ and $3 a_{2} + a_{3} = 2 a_{4}$ , then $a_{2} + a_{4} + 2 a_{5}$ is equal to ______.

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $R - (a, b)$ , then $a^{2} + b^{2}$ is equal to ......... .

If a, b and c be three distinct real numbers in G. P. and $a + b + c = x b$ , then x cannot be :

Let $3, a, b, c$ be in A.P. and $3, a - 1, b + 1, c + 9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :

Let $a_{1}, \frac{a _{2}}{2}, \frac{a _{3}}{2 ^{2}}, \dots, \frac{a _{10}}{2 ^{9}}$ be a G.P. of common ratio $\frac{1}{2}$ . If $a_{1} + a_{2} + \dots + a_{10} = 62$ , then $a_{1}$ is equal to :

Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b,c,d and p be any non zero distinct real numbers such that (a2+b2+c2)p2–2(ab+bc+cd)p+(b2+c2+d2)=0. Then :

Select Answer:

Visualized Solution

Analyzing the Given Equation

Expanding the Expression

Grouping for Perfect Squares

Rewriting as Sum of Squares

Logic of Non-negative Terms

Equating Each Term to Zero

Solving for the Common Ratio

Conclusion: Geometric Progression

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

Analyzing the Setup

The First Step

The Elegant Grouping

The Logic of Zero

The Final Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c,d and p are distinct real numbers such that (a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)≤0 then a,b,c,d

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If three distinct numbers a,b,c are in G.P. and the equations ax2+2bx+c=0 and dx2+2ex+f=0 have a common root, then which one of the following statements is correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a, b, c are in A.P. and a2,b2,c2 are in G.P. such that a<b<c and a+b+c=43​, then the value of a is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​(>0),a2​,a3​,a4​,a5​ are in a G.P., a2​+a4​=2a3​+1 and 3a2​+a3​=2a4​, then a2​+a4​+2a5​ is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a, b and c be three distinct real numbers in G. P. and a+b+c=xb, then x cannot be :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 3,a,b,c be in A.P. and 3,a−1,b+1,c+9 be in G.P. Then, the arithmetic mean of a,b and c is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,2a2​​,22a3​​,…,29a10​​ be a G.P. of common ratio 2​1​. If a1​+a2​+⋯+a10​=62, then a1​ is equal to :

Animated Solution for Mathematics - Sequence and Series: Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} -2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0$ . Then :

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0$ then $a, b, c, d$

If three distinct numbers a,b,c are in G.P. and the equations $a x^{2} + 2 b x + c = 0$ and $d x^{2} + 2 e x + f = 0$ have a common root, then which one of the following statements is correct?

If a, b, c are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. such that $a < b < c$ and $a + b + c = \frac{3}{4}$ , then the value of a is :-

If $a_{1} (> 0), a_{2}, a_{3}, a_{4}, a_{5}$ are in a G.P., $a_{2} + a_{4} = 2 a_{3} + 1$ and $3 a_{2} + a_{3} = 2 a_{4}$ , then $a_{2} + a_{4} + 2 a_{5}$ is equal to ______.

If a, b and c be three distinct real numbers in G. P. and $a + b + c = x b$ , then x cannot be :

Let $3, a, b, c$ be in A.P. and $3, a - 1, b + 1, c + 9$ be in G.P. Then, the arithmetic mean of $a, b$ and $c$ is :

Let $a_{1}, \frac{a _{2}}{2}, \frac{a _{3}}{2 ^{2}}, \dots, \frac{a _{10}}{2 ^{9}}$ be a G.P. of common ratio $\frac{1}{2}$ . If $a_{1} + a_{2} + \dots + a_{10} = 62$ , then $a_{1}$ is equal to :