Suppose $a, b, c$ are in A.P. and $a^2, 2b^2, c^2$ are in G.P. If $a < b < c$ and $a + b + c = 1$, then $9(a^2 + b^2 + c^2)$ is equal to ________ .
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Solution Steps
1
Assume terms of the A.P.
Since $a, b, c$ are in Arithmetic Progression, let:
$$a = b - d, \quad b = b, \quad c = b + d$$
Given $a < b < c$, the common difference $d$ must be greater than zero ($d > 0$).
2
Use the sum property
Given $a + b + c = 1$. Substituting the assumed terms:
$$(b - d) + b + (b + d) = 1$$
$$3b = 1 \implies b = \frac{1}{3}$$
3
Apply the G.P. condition
Given $a^2, 2b^2, c^2$ are in Geometric Progression. This implies:
$$(2b^2)^2 = a^2 \cdot c^2 \implies 4b^4 = (ac)^2$$
Taking the square root: $2b^2 = \pm ac$. Since $b^2 > 0$ and $a, c$ could have different signs, we check $ac = \pm 2b^2$.
4
Substitute $a$ and $c$ in terms of $b$ and $d$
We know $ac = (b-d)(b+d) = b^2 - d^2$.
So, $b^2 - d^2 = 2b^2$ or $b^2 - d^2 = -2b^2$.
Case A: $b^2 - d^2 = 2b^2 \implies d^2 = -b^2$ (Not possible for real $d$).
Case B: $b^2 - d^2 = -2b^2 \implies d^2 = 3b^2$.
5
Calculate $d^2$ and the terms
Since $b = 1/3$, then $d^2 = 3(1/3)^2 = 3(1/9) = 1/3$.
Now, calculate $a^2 + b^2 + c^2$:
$$a^2 + c^2 = (b-d)^2 + (b+d)^2 = 2(b^2 + d^2)$$
$$a^2 + c^2 = 2\left(\frac{1}{9} + \frac{1}{3}\right) = 2\left(\frac{1+3}{9}\right) = \frac{8}{9}$$
6
Find the final sum
$$a^2 + b^2 + c^2 = (a^2 + c^2) + b^2 = \frac{8}{9} + \frac{1}{9} = \frac{9}{9} = 1$$
The required value is $9(a^2 + b^2 + c^2) = 9(1) = 9$.
Final Answer: 9
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Theory
1. Arithmetic Progression (A.P.)
An Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference ($d$). For three terms $a, b, c$ in A.P., the middle term $b$ is the arithmetic mean of $a$ and $c$, i.e., $2b = a + c$. When the sum of three terms is given, it is mathematically convenient to assume the terms as $a-d, a, a+d$ because the $d$ cancels out when summed, directly giving the value of the middle term.
2. Geometric Progression (G.P.)
A Geometric Progression is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). For three terms $x, y, z$ to be in G.P., the square of the middle term must equal the product of the first and third terms ($y^2 = xz$). In this problem, the terms $a^2, 2b^2, c^2$ follow this rule, leading to the equation $(2b^2)^2 = a^2 c^2$.
3. Algebraic Identities in Sequences
Solving complex sequence problems often requires the use of standard algebraic identities. For terms $a = b-d$ and $c = b+d$, the product $ac$ simplifies to $b^2 - d^2$ (difference of squares). Similarly, the sum of squares $a^2 + c^2$ simplifies to $2(b^2 + d^2)$. These simplifications reduce the degree of the equations and make it easier to solve for the common difference $d$ when the middle term $b$ is already known.
4. Numerical Type Questions in JEE
Numerical value questions in JEE Main require a precise integer or decimal answer. Unlike MCQs, there is no scope for elimination or back-solving from options. Maintaining accuracy throughout calculation steps—especially with fractions and squares—is crucial. In this specific problem, the final expression $9(a^2 + b^2 + c^2)$ is designed to result in a clean integer, which is a common characteristic of JEE Main integer-type problems.
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FAQs
Q
Why did we discard d² = -b²?
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Because $a, b, c$ are real numbers, the common difference $d$ must be real. The square of a real number cannot be negative.
Q
What if we took a, a+d, a+2d?
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The answer would be the same, but the calculations would be much more tedious. Assuming $b-d, b, b+d$ is a standard trick to simplify the sum equation.
Q
Is it possible for d to be negative?
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In general A.P., yes. But here $a < b < c$ is given, which means the sequence is increasing, so $d$ must be positive.
Q
Why did we take the square root of 4b⁴ = (ac)²?
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Taking the square root simplifies the fourth-degree equation into a second-degree equation, making it easier to solve for $d^2$.
Q
What are the actual values of a and c?
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$a = 1/3 - 1/\sqrt{3}$ and $c = 1/3 + 1/\sqrt{3}$.
Q
Can a², 2b², c² be zero?
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Since $b=1/3$, $2b^2 \neq 0$. In a G.P., if the middle term is non-zero, the other terms must also be non-zero.
Q
Does 9(a² + b² + c²) always result in an integer?
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Not necessarily in every problem, but in JEE integer-type questions, the expression is usually tailored to yield a clean whole number.
Q
Is there any other value for d²?
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If we had $ac = 2b^2$, we get $d^2 = -b^2$. If we use $ac = -2b^2$, we get $d^2 = 3b^2$. Only the latter gives real values.
Q
What is the common ratio of the G.P.?
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The common ratio $r = 2b^2 / a^2$. Using $d^2=3b^2$, $a^2 = (b-b\sqrt{3})^2 = b^2(4-2\sqrt{3})$. Ratio $r = 2/(4-2\sqrt{3}) = 1/(2-\sqrt{3}) = 2+\sqrt{3}$.
Q
How to verify the answer?
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Check if $a+b+c=1$ and if $(2b^2)^2 = a^2c^2$ with $b=1/3$ and $d=1/\sqrt{3}$. Both conditions are satisfied.
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