If a, b, c are in A.P. and a², b², c² are in G.P. such that a

Q: Wait, why can't I just say $b^2 = ac$? Isn't that the definition of G.P.?

That is a classic trap! That formula applies only if $a, b, c$ are in G.P. Here, the problem explicitly states that $a^2, b^2, c^2$ are in G.P. Therefore, the middle term of *that* sequence is $b^2$. So, the condition becomes $(b^2)^2 = a^2 c^2$. Always read the sequence terms carefully!

Q: Why did we reject $d=0$ in the first case?

Great observation! The problem states $a < b < c$. If $d=0$, then $a = b = c = \frac{1}{4}$. This would make the sequence constant, not strictly increasing. In JEE problems, always check your constraints against your solutions!

Q: Why do we need the modulus when taking the square root of $b^4 = (ac)^2$?

Because $\sqrt{x^2}$ is always $|x|$. If you ignore the modulus, you miss the second case where the product $ac$ is negative. In competitive exams, missing that modulus is often the difference between getting the right answer and falling into a distractor option.

Question

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 :-

Options:

(A) $\frac{1}{4} - \frac{1}{\sqrt{2}}$

(B) $\frac{1}{4} - \frac{1}{3\sqrt{2}}$

(C) $\frac{1}{4} - \frac{1}{4\sqrt{2}}$

(D) $\frac{1}{4} - \frac{1}{2\sqrt{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup We are given three numbers $a, b, c$ that form an Arithmetic Progression (A.P.). In any A.P., the middle term is the arithmetic mean of its neighbors, which gives us the symmetry condition: $$2b = a + c$$ We are also provided with the sum constraint $a + b + c = \frac{3}{4}$. By substituting the symmetry condition into this sum, we obtain: $$(a + c) + b = \frac{3}{4} \Rightarrow 2b + b = \frac{3}{4}$$ Solving this yields $3b = \frac{3}{4}$, which fixes the middle term at **$b = \frac{1}{4}$**. ### The Geometric Constraint The problem states that $a^2, b^2, c^2$ are in a Geometric Progression (G.P.). The condition for three terms to be in G.P. is that the square of the middle term equals the product of the extremes: $$(b^2)^2 = a^2 c^2$$ This simplifies to $b^4 = (ac)^2$. Taking the square root of both sides, we must account for the absolute value: $$b^2 = |ac|$$ ### Solving for the Sequence Let $d$ be the common difference of the A.P. Since $a < b < c$, we define the terms as $a = \frac{1}{4} - d$ and $c = \frac{1}{4} + d$, where $d > 0$. Substituting these into our G.P. condition: $$\left( \frac{1}{4} \right)^2 = \left| \left( \frac{1}{4} - d \right) \left( \frac{1}{4} + d \right) \right|$$ Applying the difference of squares identity, we get: $$\frac{1}{16} = \left| \frac{1}{16} - d^2 \right|$$ ### Final Calculation We evaluate the two possible cases for the absolute value: **Case 1:** $\frac{1}{16} - d^2 = \frac{1}{16} \Rightarrow d^2 = 0$. We reject this as the sequence must be strictly increasing ($d > 0$). **Case 2:** $\frac{1}{16} - d^2 = -\frac{1}{16} \Rightarrow d^2 = \frac{2}{16} = \frac{1}{8}$. Thus, $d = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$. Finally, we calculate the value of $a$: $$a = b - d = \frac{1}{4} - \frac{1}{2\sqrt{2}}$$ The value of $a$ is **$\frac{1}{4} - \frac{1}{2\sqrt{2}}$**.

Animated Solution for Mathematics - Sequence and Series: 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 :-

Select Answer:

Visualized Solution

Problem Setup

A.P. Condition

Using the Sum

Solving for $b$

Common Difference $d$

G.P. Condition

Simplifying G.P. Equation

Substituting Values

Difference of Squares

Case 1: Positive Argument

Case 2: Negative Argument

Value of $d$

Calculating $a$

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

Analyzing the Setup

The Geometric Constraint

Solving for the Sequence

Final Calculation

Similar Questions

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 :

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 $a, b$ and $c$ be in G. P. with common ratio $r$ , where $a \neq = 0$ and $0 < r \leq 1/2$ . If $3 a, 7 b$ and $15 c$ are the first three terms of an A. P., then the 4th term of this A. P. 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 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?

Let $a, b$ and $c$ be the 7th, 11th and 13th terms respectively of a non-constant A.P. If these are also the three consecutive terms of a G.P., then $a / c$ is equal to:

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

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 ......... .

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:

Three numbers are in an increasing geometric progression with common ratio $r$ . If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference $d$ . If the fourth term of $GP$ is $3 r^{2}$ , then $r^{2} - d$ 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; } 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 :-