Why can we replace with ?

We are given that are in G.P., so by definition , which gives .

JEE MainMathematics

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Sequences and Series: JEE Main Mathematics Question with Solution

Q: Why do the two equations share a root just because the parabolas intersect at y = 1?

An intersection point satisfies both curves. When , the -coordinate must simultaneously satisfy both quadratic equations in .

Q: Why wasn't testing sufficient?

Because when , , , and , making option (1) and option (2) identical. A non-trivial common ratio like breaks the degeneracy.

The parabolas :

\mathrm{ax}^2 + 2\mathrm{bx} + \mathrm{cy} = 0

and

\mathrm{dx}^2 + 2\mathrm{ex} + \mathrm{fy} = 0

intersect on the line

\mathrm{y} = 1

. If

\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}

are positive real numbers and

\mathrm{a}, \mathrm{b}, \mathrm{c}

are in G.P., then

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\mathrm{d}, \mathrm{e}, \mathrm{f}

are in G.P.B

\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}

are in A.P.C

\mathrm{d}, \mathrm{e}, \mathrm{f}

are in A.P.D

\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}

are in G.P.

Step-by-step solutionView

Correct answer

Setting $y = 1$ gives $ax^2 + 2bx + c = 0$ with equal roots $x = -b/a$ since $b^2 = ac$ ; substituting this root into $dx^2 + 2ex + f = 0$ reveals that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ form an arithmetic progression.

Option analysis

Why each option works or fails

A ·

\mathrm{d}, \mathrm{e}, \mathrm{f}

are in G.P.

Assuming that a geometric progression in the first parabola's coefficients directly induces a geometric progression in the second parabola's standalone coefficients. Substitute the common root $x = -b/a$ into the second equation and divide by $c$ to see that the ratio involves terms of the form $d/a$ , $e/b$ , and $f/c$ , not isolated coefficients.

B ·

\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}

are in A.P.

None. This is the correct option. Correct: At $y = 1$ , $ax^2 + 2bx + c = 0$ has discriminant $4(b^2 - ac) = 0$ , giving a repeated root $x = -b/a$ . Since both parabolas intersect on $y = 1$ , $x = -b/a$ must satisfy $dx^2 + 2ex + f = 0$ . Substituting gives $d(b^2/a^2) - 2e(b/a) + f = 0$ . Using $b^2 = ac$ , this simplifies to $\frac{d}{a} - 2\frac{e}{b} + \frac{f}{c} = 0$ , which proves that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

C ·

\mathrm{d}, \mathrm{e}, \mathrm{f}

are in A.P.

Ignoring the scaling weights $a, b, c$ when evaluating the linear combination of $d, e, f$ . Remember that the coefficients of the second parabola are coupled to the roots through $b/a$ , so the relation must involve the ratios $d/a, e/b, f/c$ rather than $d, e, f$ directly.

D ·

\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}

are in G.P.

Confusing the condition $2(e/b) = d/a + f/c$ with a geometric progression condition due to misinterpreting the middle term's multiplier of $2$ . A relation of the form $2Y = X + Z$ characterizes an arithmetic progression, not a geometric progression.

Reviewed route

Solution

StepWorking

01given

Parabolas $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on $y = 1$ . The coefficients $a, b, c, d, e, f$ are positive real numbers, and $a, b, c$ are in G.P. ( $b^2 = ac$ ).

02goal

Determine the progression relation among $d, e, f$ or $d/a, e/b, f/c$ .

03approach

Substitute $y = 1$ into both equations to get quadratic equations in $x$ sharing a real root. Since $b^2 = ac$ , the first quadratic has discriminant zero and thus a unique repeated root. Substitute this root into the second equation and divide by $c$ to establish the arithmetic progression.

04execute

Substituting $y = 1$ into both equations yields $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ . For the first equation, discriminant $D_1 = (2b)^2 - 4ac = 4(b^2 - ac) = 0$ since $a, b, c$ are in G.P. Hence, the repeated root is $x = -2b / (2a) = -b/a$ .

05execute

Since the curves intersect on $y = 1$ , $x = -b/a$ must satisfy $dx^2 + 2ex + f = 0$ : $d(-b/a)^2 + 2e(-b/a) + f = 0 \implies d(b^2/a^2) - 2eb/a + f = 0$ . Using $b^2 = ac$ , this becomes $d(ac/a^2) - 2eb/a + f = 0 \implies (dc/a) - 2eb/a + f = 0$ . Dividing the entire equation by $c$ : $d/a - 2eb/(ac) + f/c = 0$ . Since $ac = b^2$ , $2eb/(ac) = 2eb/b^2 = 2e/b$ . Thus, $d/a + f/c = 2(e/b)$ , which implies $d/a, e/b, f/c$ are in A.P.

✓verify

Let $a=1, b=2, c=4$ (in G.P.). The first equation at $y=1$ is $x^2 + 4x + 4 = (x+2)^2 = 0$ , giving $x = -2$ . For $d/a, e/b, f/c$ to be in A.P., choose $d/a = 1 \implies d = 1$ , $e/b = 2 \implies e = 4$ , $f/c = 3 \implies f = 12$ . The second equation at $y=1$ is $1(-2)^2 + 2(4)(-2) + 12 = 4 - 16 + 12 = 0$ , which holds identically.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do the two equations share a root just because the parabolas intersect at y = 1?

An intersection point $(x_0, y_0)$ satisfies both curves. When $y_0 = 1$ , the $x$ -coordinate $x_0$ must simultaneously satisfy both quadratic equations in $x$ .

Why can we replace

ac

with

b^2

We are given that $a, b, c$ are in G.P., so by definition $b/a = c/b$ , which gives $b^2 = ac$ .

Why wasn't testing

a=b=c=1

sufficient?

Because when $a=b=c=1$ , $d/a = d$ , $e/b = e$ , and $f/c = f$ , making option (1) and option (2) identical. A non-trivial common ratio like $r=2$ breaks the degeneracy.

Answer

Setting $y = 1$ gives $ax^2 + 2bx + c = 0$ with equal roots $x = -b/a$ since $b^2 = ac$ ; substituting this root into $dx^2 + 2ex + f = 0$ reveals that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ form an arithmetic progression.

Why each option works or fails

A: $\mathrm{d}, \mathrm{e}, \mathrm{f}$ are in G.P. - Assuming that a geometric progression in the first parabola's coefficients directly induces a geometric progression in the second parabola's standalone coefficients. Substitute the common root $x = -b/a$ into the second equation and divide by $c$ to see that the ratio involves terms of the form $d/a$ , $e/b$ , and $f/c$ , not isolated coefficients.
B · correct: $\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}$ are in A.P. - None. This is the correct option. Correct: At $y = 1$ , $ax^2 + 2bx + c = 0$ has discriminant $4(b^2 - ac) = 0$ , giving a repeated root $x = -b/a$ . Since both parabolas intersect on $y = 1$ , $x = -b/a$ must satisfy $dx^2 + 2ex + f = 0$ . Substituting gives $d(b^2/a^2) - 2e(b/a) + f = 0$ . Using $b^2 = ac$ , this simplifies to $\frac{d}{a} - 2\frac{e}{b} + \frac{f}{c} = 0$ , which proves that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.
C: $\mathrm{d}, \mathrm{e}, \mathrm{f}$ are in A.P. - Ignoring the scaling weights $a, b, c$ when evaluating the linear combination of $d, e, f$ . Remember that the coefficients of the second parabola are coupled to the roots through $b/a$ , so the relation must involve the ratios $d/a, e/b, f/c$ rather than $d, e, f$ directly.
D: $\frac{\mathrm{d}}{\mathrm{a}}, \frac{\mathrm{e}}{\mathrm{b}}, \frac{\mathrm{f}}{\mathrm{c}}$ are in G.P. - Confusing the condition $2(e/b) = d/a + f/c$ with a geometric progression condition due to misinterpreting the middle term's multiplier of $2$ . A relation of the form $2Y = X + Z$ characterizes an arithmetic progression, not a geometric progression.

Step-by-step solution

given: Parabolas $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on $y = 1$ . The coefficients $a, b, c, d, e, f$ are positive real numbers, and $a, b, c$ are in G.P. ( $b^2 = ac$ ).
goal: Determine the progression relation among $d, e, f$ or $d/a, e/b, f/c$ .
approach: Substitute $y = 1$ into both equations to get quadratic equations in $x$ sharing a real root. Since $b^2 = ac$ , the first quadratic has discriminant zero and thus a unique repeated root. Substitute this root into the second equation and divide by $c$ to establish the arithmetic progression.
execute: Substituting $y = 1$ into both equations yields $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ . For the first equation, discriminant $D_1 = (2b)^2 - 4ac = 4(b^2 - ac) = 0$ since $a, b, c$ are in G.P. Hence, the repeated root is $x = -2b / (2a) = -b/a$ .
execute: Since the curves intersect on $y = 1$ , $x = -b/a$ must satisfy $dx^2 + 2ex + f = 0$ : $d(-b/a)^2 + 2e(-b/a) + f = 0 \implies d(b^2/a^2) - 2eb/a + f = 0$ . Using $b^2 = ac$ , this becomes $d(ac/a^2) - 2eb/a + f = 0 \implies (dc/a) - 2eb/a + f = 0$ . Dividing the entire equation by $c$ : $d/a - 2eb/(ac) + f/c = 0$ . Since $ac = b^2$ , $2eb/(ac) = 2eb/b^2 = 2e/b$ . Thus, $d/a + f/c = 2(e/b)$ , which implies $d/a, e/b, f/c$ are in A.P.
verify: Let $a=1, b=2, c=4$ (in G.P.). The first equation at $y=1$ is $x^2 + 4x + 4 = (x+2)^2 = 0$ , giving $x = -2$ . For $d/a, e/b, f/c$ to be in A.P., choose $d/a = 1 \implies d = 1$ , $e/b = 2 \implies e = 4$ , $f/c = 3 \implies f = 12$ . The second equation at $y=1$ is $1(-2)^2 + 2(4)(-2) + 12 = 4 - 16 + 12 = 0$ , which holds identically.

Shortcut:

When to use it: When the relation is invariant under choice of valid progressions, test with simple integers.

given: $a, b, c$ in G.P., so pick $a = 1, b = 1, c = 1$ .

goal: Find the progression satisfied by $d, e, f$ and $d/a, e/b, f/c$ under this assignment.

approach: With $a=b=c=1$ , the first equation at $y=1$ is $x^2 + 2x + 1 = 0 \implies x = -1$ . The second equation becomes $d(-1)^2 + 2e(-1) + f = 0 \implies d - 2e + f = 0$ . Check which option this matches.

execute: For $a=b=c=1$ , $d - 2e + f = 0$ means $2e = d + f$ , so $d, e, f$ are in A.P., and also $d/a, e/b, f/c$ are in A.P. (since $a=b=c=1$ ). Now pick $a=1, b=2, c=4$ : root is $x = -2$ . The second equation is $4d - 4e + f = 0$ , i.e., $d/1 - 2(e/2) + f/4 = 0 \implies d/a + f/c = 2(e/b)$ , confirming $d/a, e/b, f/c$ are in A.P. while $d, e, f$ are not in general in A.P. (e.g., $d=1, e=2 \implies f=4$ , where $d, e, f$ are $1, 2, 4$ , not in A.P.).

verify: Only option (1) holds across both choices of $(a, b, c)$ .

Sequences and Series: JEE Main Mathematics Question with Solution

What feels right?

Setting $y = 1$ gives $ax^2 + 2bx + c = 0$ with equal roots $x = -b/a$ since $b^2 = ac$ ; substituting this root into $dx^2 + 2ex + f = 0$ reveals that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ form an arithmetic progression.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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