Q.If $a, b, c, d$ and $p$ are different real numbers such that $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \leq 0$, then show that $a, b, c$ and $d$ are in G.P.

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Consider the following Assertion (A): $\dfrac{1}{5 \cdot 9} + \dfrac{1}{9 \cdot 13} + \dfrac{1}{13 \cdot 17} + \ldots$ to 10 terms $= \dfrac{9}{41}$ Reason (R): For all $n \in \mathbb{N}$, $\dfrac{1}{5 \cdot 9} + \dfrac{1}{9 \cdot 13} + \dfrac{1}{13 \cdot 17} + \ldots$ to $n$ terms $= \dfrac{n}{5(4n+5)}$ The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   The given series is a telescoping series. By finding its general sum $S_n$, we find that Reason (R) is true, but Assertion (A) is false. The correct option is (D).

   The problem asks us to evaluate an Assertion (A) and a Reason (R) related to the sum of a specific series. To do this, we need to find a general formula for the sum of the first $n$ terms of the series. The series is of a type that can often be summed using the method of differences, specifically by expressing each term as a difference of two quantities, leading to a telescoping sum.

   Concept: Telescoping Series and Partial Fractions

   A telescoping series is a series where most of the terms cancel out, leaving only the first and last terms (or a few terms at the beginning and end). This cancellation typically happens when each term can be expressed as a difference of two consecutive terms of a sequence. For a series of the form $\sum_{k=1}^{n} (f(k) - f(k+c))$, where $c$ is a constant, many intermediate terms will cancel out.

   In this problem, each term of the series is a fraction with a product of two factors in the denominator. This structure suggests using partial fraction decomposition to rewrite each term as a difference.

   1. Identify the general term of the series: The given series is $\dfrac{1}{5 \cdot 9} + \dfrac{1}{9 \cdot 13} + \dfrac{1}{13 \cdot 17} + \ldots$. Let's look at the denominators: $5 \cdot 9$, $9 \cdot 13$, $13 \cdot 17$. The first factor in each product forms an arithmetic progression: $5, 9, 13, \ldots$. The common difference is $9-5=4$. So, the $k$-th term of this sequence is $5 + (k-1)4 = 4k+1$. The second factor in each product forms an arithmetic progression: $9, 13, 17, \ldots$. The common difference is $13-9=4$. So, the $k$-th term of this sequence is $9 + (k-1)4 = 4k+5$. Thus, the general $k$-th term of the series, $T_k$, is given by:

   $T_k = \frac{1}{(4k+1)(4k+5)}$

   2. Express the general term using partial fractions: We want to write $T_k$ in the form $\dfrac{A}{4k+1} + \dfrac{B}{4k+5}$.

   $\frac{1}{(4k+1)(4k+5)} = \frac{A}{4k+1} + \frac{B}{4k+5}$

   Multiplying both sides by $(4k+1)(4k+5)$:
   

   $1 = A(4k+5) + B(4k+1)$

   To find $A$, set $4k+1=0 \implies k = -1/4$:
   

   $1 = A(4(-1/4)+5) + B(0) \implies 1 = A(4) \implies A = \frac{1}{4}$

   To find $B$, set $4k+5=0 \implies k = -5/4$:
   

   $1 = A(0) + B(4(-5/4)+1) \implies 1 = B(-4) \implies B = -\frac{1}{4}$

   So, the general term $T_k$ can be written as:
   

   $T_k = \frac{1}{4} \left( \frac{1}{4k+1} - \frac{1}{4k+5} \right)$

   > [!TIP]
   > For terms of the form $\frac{1}{a \cdot b}$ where $b-a=d$, we can often write $\frac{1}{a \cdot b} = \frac{1}{d} \left( \frac{1}{a} - \frac{1}{b} \right)$. Here, $a = 4k+1$ and $b = 4k+5$, so $d = (4k+5) - (4k+1) = 4$. This directly gives $T_k = \frac{1}{4} \left( \frac{1}{4k+1} - \frac{1}{4k+5} \right)$.
   

   3. Calculate the sum of the first $n$ terms ($S_n$):

   The sum $S_n$ is $\sum_{k=1}^{n} T_k$:

   $S_n = \sum_{k=1}^{n} \frac{1}{4} \left( \frac{1}{4k+1} - \frac{1}{4k+5} \right)$

   $S_n = \frac{1}{4} \left[ \left( \frac{1}{4(1)+1} - \frac{1}{4(1)+5} \right) + \left( \frac{1}{4(2)+1} - \frac{1}{4(2)+5} \right) + \ldots + \left( \frac{1}{4n+1} - \frac{1}{4n+5} \right) \right]$

   $S_n = \frac{1}{4} \left[ \left( \frac{1}{5} - \frac{1}{9} \right) + \left( \frac{1}{9} - \frac{1}{13} \right) + \left( \frac{1}{13} - \frac{1}{17} \right) + \ldots + \left( \frac{1}{4n+1} - \frac{1}{4n+5} \right) \right]$

   This is a telescoping sum. The intermediate terms cancel out: $-\frac{1}{9}$ cancels with $+\frac{1}{9}$, $-\frac{1}{13}$ cancels with $+\frac{1}{13}$, and so on. …
   

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2. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If the difference of the roots of the equation $x^2 - 7x + 10 = 0$ is same as the difference of the roots of the equation $x^2 - 17x + k = 0$, then a divisor of k is (A) $14$ (B) $17$ (C) $6$ (D) $15$

   ›Reveal solutionSolution

   Matching the "difference of roots" of the two quadratics gives $k=70$; among the given options only $14$ divides it.

   Concept and Intuition

   For a quadratic $x^2-(\text{sum})x+(\text{product})=0$, the difference of its two roots can be found without solving for the roots individually, using $(\text{root}_1-\text{root}_2)^2 = (\text{sum})^2-4(\text{product})$ — this follows directly from $(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2$.

   Step-by-Step Solution

   1. For $x^2-7x+10=0$: roots are $2$ and $5$ (factors as $(x-2)(x-5)$), so the difference of roots is $5-2=3$.
   2. For $x^2-17x+k=0$: sum of roots $=17$, product $=k$. Difference$^2 = 17^2-4k = 289-4k$.
   3. Set the two differences equal (matching magnitudes): $289-4k = 3^2 = 9$.
   4. Solve: $4k = 289-9=280 \Rightarrow k=70$. …

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3. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If the coefficient of $x^r$ in the expansion of $(1 + x + x^2 + x^3)^{100}$ is $a_r$ and $S = \sum_{r=0}^{300} a_r$, then $\sum_{r=0}^{300} r a_r =$ (A) $(50) S$ (B) $(25) S$ (C) $(150) S$ (D) $(100) S$

   ›Reveal solutionSolution

   The sum $\sum r a_r$ is the derivative of the generating function evaluated at $x=1$, which equals $100 \cdot (1+1+1+1)^{99} \cdot (1+2+3) = 100 \cdot 4^{99} \cdot 6$, while $S = 4^{100}$, so the ratio is $150$, making the answer $150S$, option (C).

   We are given the expansion of $(1 + x + x^2 + x^3)^{100}$. Let $a_r$ be the coefficient of $x^r$, and $S = \sum_{r=0}^{300} a_r$. We need $\sum_{r=0}^{300} r a_r$.

   Concept and intuition:

   The sum $\sum r a_r$ is like a weighted sum where each coefficient is multiplied by its exponent. This is exactly what you get if you differentiate the polynomial (or series) and then set $x=1$. Why? Because if $P(x) = \sum a_r x^r$, then $P'(x) = \sum r a_r x^{r-1}$, so $P'(1) = \sum r a_r$. Meanwhile, $S = P(1)$. So the problem reduces to computing $P(1)$ and $P'(1)$ for $P(x) = (1 + x + x^2 + x^3)^{100}$, then finding the ratio.

   Step-by-step:

   1. Define the generating function.

      Let $P(x) = (1 + x + x^2 + x^3)^{100}$. Then $P(x) = \sum_{r=0}^{300} a_r x^r$.

      So $S = P(1)$ and $\sum r a_r = P'(1)$.

   2. Compute $S = P(1)$.

      At $x=1$, each term $1 + x + x^2 + x^3$ becomes $1+1+1+1 = 4$.

      Hence $P(1) = 4^{100}$. So $S = 4^{100}$.

   3. Compute $P'(x)$ using the chain rule.

      Write $P(x) = [f(x)]^{100}$ where $f(x) = 1 + x + x^2 + x^3$.

      Then $P'(x) = 100 [f(x)]^{99} \cdot f'(x)$.

      Here $f'(x) = 1 + 2x + 3x^2$.

   4. Evaluate $P'(1)$.

      At $x=1$:

      $f(1) = 4$, so $[f(1)]^{99} = 4^{99}$.

      $f'(1) = 1 + 2 + 3 = 6$.

      Therefore $P'(1) = 100 \cdot 4^{99} \cdot 6 = 600 \cdot 4^{99}$.

   5. Find the ratio $\frac{P'(1)}{P(1)}$.

      $$$$

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4. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If the roots of $\sqrt{\dfrac{1-y}{y}} + \sqrt{\dfrac{y}{1-y}} = \dfrac{5}{2}$ are $\alpha$ and $\beta$ ($\beta > \alpha$) and the equation $(\alpha+\beta)x^4 - 25\alpha\beta x^2 + (\gamma + \beta - \alpha) = 0$ has real roots, then a possible value of $\gamma$ is (A) $\dfrac{1}{2}$ (B) $4$ (C) $2\pi$ (D) $\sqrt{e + 13}$

   ›Reveal solutionSolution

   Solve the surd equation via a reciprocal substitution to get $\alpha,\beta$, then find which $\gamma$ keeps the resulting biquadratic real-rooted. Answer: $\gamma=1/2$.

   Concept and Intuition

   An equation of the form $\sqrt{u}+\dfrac1{\sqrt u}=k$ (here $u=y/(1-y)$) is best handled by substituting $t=\sqrt u$, turning it into a simple quadratic $t^2-kt+1=0$. Once $\alpha,\beta$ are known, the quartic in $x$ becomes a quadratic in $u=x^2$, and "real roots in $x$" needs a non-negative discriminant AND a non-negative root $u$.

   Step-by-Step Solution

   1. Let $t=\sqrt{y/(1-y)}$, so the equation is $t+1/t=5/2 \Rightarrow 2t^2-5t+2=0 \Rightarrow (2t-1)(t-2)=0 \Rightarrow t=2,\,1/2$.
   2. $y=t^2/(1+t^2)$: for $t=2$, $y=4/5$; for $t=1/2$, $y=1/5$. So $\alpha=1/5,\ \beta=4/5$.
   3. $\alpha+\beta=1$, $\alpha\beta=4/25$, $\beta-\alpha=3/5$.
   4. Quartic: $x^4-4x^2+(\gamma+3/5)=0$. Let $u=x^2$: $u^2-4u+(\gamma+3/5)=0$. …

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5. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If 'a' is a rational number, then the roots of the equation $x^2-3ax+a^2-2a-4=0$ are (A) rational and equal numbers (B) different real numbers (C) different rational numbers only (D) not real numbers

   ›Reveal solutionSolution

   This tests analyzing a discriminant that itself is a quadratic in a parameter; the roots are always real and different, but not always rational — the answer is (B).

   Concept and Intuition

   For $x^2-3ax+(a^2-2a-4)=0$, whether the roots are real/equal/rational depends entirely on the discriminant $D=9a^2-4(a^2-2a-4)$. Since $a$ itself varies over all rationals, we must check whether $D$ is always positive (real & distinct roots guaranteed) and separately whether $D$ is always a perfect square (rational roots guaranteed) — these are two independent questions.

   Step-by-Step Solution

   1. Compute $D = 9a^2-4(a^2-2a-4) = 9a^2-4a^2+8a+16 = 5a^2+8a+16$.
   2. Treat $D$ as a quadratic in $a$: its own discriminant is $8^2-4\cdot5\cdot16 = 64-320=-256<0$, and the leading coefficient $5>0$. So $D(a)>0$ for every real value of $a$ — meaning the roots of the original quadratic are always real and always distinct (never equal), regardless of which rational $a$ is chosen.
   3. Check whether $D$ is always a perfect square for rational $a$: at $a=0$, $D=16=4^2$ (rational roots); but at $a=1$, $D=5+8+16=29$, which is not a perfect square, so $\sqrt{D}$ is irrational and the roots are irrational real numbers. …

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6. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If $P$ is a non-singular matrix such that $I + P + P^2 + \ldots + P^n = O$ ($O$ denotes the null matrix), then $P^{-1} =$ (A) $P^n$ (B) $-P^n$ (C) $-(I + P + \ldots + P^n)$ (D) $-(I + P + \ldots + P^{n-1})$

   ›Reveal solutionSolution

   Multiplying $I+P+\dots+P^n=O$ by $P^{-1}$ and then folding the original relation back in gives $P^{-1}=P^{n}$ — a strikingly clean result. The correct choice is option (A).

   The concept first

   Matrices don't allow division, but a non-singular matrix behaves like a nonzero number in one crucial way: you may multiply an equation through by its inverse. The whole trick here is to notice that the series is finite and sums to the null matrix, which pins the top power $P^n$ against everything below it.

   Step-by-step

   Step 1 — Write the hypothesis.

   $I + P + P^2 + \dots + P^{n-1} + P^{n} = O$

   Step 2 — Isolate the top power.

   $I + P + \dots + P^{n-1} = -P^{n} \qquad (\star)$

   Step 3 — Multiply the hypothesis on the left by $P^{-1}$ (legal because $P$ is non-singular):

   $P^{-1}I + P^{-1}P + P^{-1}P^2 + \dots + P^{-1}P^{n} = P^{-1}O = O$

   $\Rightarrow\ P^{-1} + I + P + \dots + P^{n-1} = O$

   Step 4 — Substitute $(\star)$. …

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7. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If the values of k for which the equation $x^2+2(k+2)x+6k+7=0$ has equal roots are $k_1$ and $k_2$, then $k_1^2+k_2^2=$ (A) 8 (B) 9 (C) 10 (D) 12

   ›Reveal solutionSolution

   The condition for equal roots is discriminant $=0$; solving the resulting quadratic in $k$ gives $k=3$ and $k=-1$, so $k_1^2+k_2^2=10$.

   Concept and Intuition

   A quadratic $ax^2+bx+c=0$ has equal (repeated) roots exactly when its discriminant $b^2-4ac=0$. Here the coefficients themselves depend on a parameter $k$, so setting the discriminant to zero produces a new quadratic equation whose roots are the two values of $k$ that make the original equation have a double root.

   Step-by-Step Solution

   1. For $x^2+2(k+2)x+(6k+7)=0$: $a=1,\ b=2(k+2),\ c=6k+7$.
   2. Discriminant $=0$: $[2(k+2)]^2-4(1)(6k+7)=0 \Rightarrow 4(k+2)^2-4(6k+7)=0$.
   3. Divide by 4: $(k+2)^2-(6k+7)=0 \Rightarrow k^2+4k+4-6k-7=0 \Rightarrow k^2-2k-3=0$. …

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8. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If the roots of the equation $3x^2+4kx+3=0$ are non-real, then k lies in the interval (A) $\left[-2, \dfrac{-3}{2}\right]$ (B) $\left[\dfrac{3}{2}, 2\right]$ (C) $\left(\dfrac{-3}{2}, \dfrac{3}{2}\right)$ (D) $(2, 3)$

   ›Reveal solutionSolution

   Non-real roots of a quadratic require a strictly negative discriminant; solving that inequality for $k$ gives the open interval $(-3/2, 3/2)$.

   Concept and Intuition

   For $ax^2+bx+c=0$ with real coefficients, the roots are non-real exactly when the discriminant $b^2-4ac<0$ (strictly, since $=0$ gives a repeated real root and $>0$ gives two distinct real roots).

   Step-by-Step Solution

   1. Here $a=3,\ b=4k,\ c=3$.
   2. Discriminant $= (4k)^2-4(3)(3) = 16k^2-36$.
   3. Require $16k^2-36<0 \Rightarrow 16k^2<36 \Rightarrow k^2<\dfrac{36}{16}=\dfrac{9}{4}$.
   4. Taking square roots: $|k|<\dfrac{3}{2}$, i.e. $k\in\left(-\dfrac{3}{2},\dfrac{3}{2}\right)$. …

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9. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If the difference between the roots of $x^2+ax+b=0$ and that of the roots of $x^2+bx+a=0$ is same and $a \neq b$, then (A) $a-b-4=0$ (B) $a-b+4=0$ (C) $a+b+4=0$ (D) $a+b-4=0$

   ›Reveal solutionSolution

   Equating the (squared) root-differences of the two quadratics and factoring out $(a-b)$ (which is nonzero) leaves $a+b+4=0$.

   Concept and Intuition

   For a monic quadratic $x^2+px+q=0$, if the roots are $\alpha,\beta$, then $\alpha-\beta = \pm\sqrt{(\alpha+\beta)^2-4\alpha\beta} = \pm\sqrt{p^2-4q}$ (using $\alpha+\beta=-p,\ \alpha\beta=q$). So the magnitude of the difference of roots depends only on $\sqrt{p^2-4q}$.

   Step-by-Step Solution

   1. For $x^2+ax+b=0$: difference of roots (in magnitude) $=\sqrt{a^2-4b}$.
   2. For $x^2+bx+a=0$: difference of roots (in magnitude) $=\sqrt{b^2-4a}$.
   3. Given these are equal: $\sqrt{a^2-4b} = \sqrt{b^2-4a}$. Squaring (valid since both sides are non-negative real quantities under the given condition): $a^2-4b = b^2-4a$.
   4. Rearranging: $a^2-b^2 -4b+4a = 0 \Rightarrow (a-b)(a+b) + 4(a-b) = 0 \Rightarrow (a-b)(a+b+4)=0$. …

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10. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.For what values of $a \in Z$, the quadratic expression $(x+a)(x+1991)+1$ can be factorised as $(x+b)(x+c)$, where $b,c \in Z$? (A) $1990$ (B) $1989$ (C) $1991$ (D) $1992$

    ›Reveal solutionSolution

    $(x+a)(x+1991)+1$ becomes a perfect square $(x+m)^2$ exactly when $a$ and $1991$ are $2$ apart (symmetric about the same midpoint $m$); among the choices this happens at $a=1989$.

    Concept and Intuition

    For two numbers $p,q$ with mean $m=\frac{p+q}{2}$, we can write $p=m-d$, $q=m+d$ where $d=\frac{q-p}{2}$. Then $(x+p)(x+q) = (x+m-d)(x+m+d) = (x+m)^2-d^2$. Adding $1$ gives $(x+m)^2 - d^2+1$, which becomes a perfect square $(x+m)^2$ precisely when $d^2=1$, i.e. $d=\pm1$ — meaning $p$ and $q$ differ by exactly $2$.

    Step-by-Step Solution

    1. Here $q=1991$ (fixed) and $p=a$ (to determine), so we want $|a-1991|=2$, i.e. $a=1989$ or $a=1993$.
    2. Check $a=1989$: midpoint $m=\frac{1989+1991}{2}=1990$, $d=1$. So $(x+1989)(x+1991)+1=(x+1990)^2-1+1=(x+1990)^2$ — a perfect square with $b=c=1990$, both integers. ✓
    3. Check the other listed options directly using $(b-c)^2=(b+c)^2-4bc$ (must be a non-negative perfect square for integer $b,c$ to exist):
       • $a=1990$: $b+c=3981$, $bc=1990\cdot1991+1$; $(b-c)^2 = (1991-1990)^2-4=1-4=-3 <0$ — no real (hence no integer) solution.
       • $a=1991$: gives $(x+1991)^2+1$, and $(b-c)^2=-4<0$ — no solution. …

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11. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.The number of real values of m so that the equation $x^2 + (2m+1)x + m = 0$ has equal roots is (A) 1 (B) 0 (C) 2 (D) 3

    ›Reveal solutionSolution

    Setting the discriminant of the quadratic to zero (the condition for equal/repeated roots) leads to an equation with no real solution, so no real $m$ works.

    Concept and Intuition

    A quadratic $ax^2+bx+c=0$ has equal (repeated) roots exactly when its discriminant $b^2-4ac=0$. Here $a=1,\ b=2m+1,\ c=m$, so we set up and solve this discriminant condition for $m$.

    Step-by-Step Solution

    1. Discriminant: $D=(2m+1)^2-4(1)(m)=4m^2+4m+1-4m=4m^2+1$.
    2. Set $D=0$: $4m^2+1=0\Rightarrow m^2=-\dfrac14$.
    3. Since $m^2$ cannot be negative for real $m$, there is no real value of $m$ satisfying this equation. …

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12. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.If a, b, c, d are real numbers such that $a<b<c<d$, then the roots of the equation $(x-a)(x-c)+2(x-b)(x-d)=0$ are (A) Real & need not be distinct (B) Real and distinct (C) Non-real and distinct (D) Non-real and need not be distinct

    ›Reveal solutionSolution

    This tests root-location by sign analysis (intermediate value theorem) rather than solving the quadratic explicitly. The roots are real and distinct.

    Concept and Intuition

    When a quadratic is built as a sum of two products like $(x-a)(x-c)+k(x-b)(x-d)$ with $k>0$, you don't need to expand and use the discriminant. Instead, evaluate the quadratic at the four given points $a,b,c,d$. Wherever the sign flips between consecutive points, a root must lie strictly in between (continuity + IVT). Since a quadratic has at most 2 roots, finding 2 sign changes locates BOTH roots precisely, and they must be real and distinct because they sit in two separate, non-overlapping open intervals.

    Step-by-Step Solution

    1. Let $f(x)=(x-a)(x-c)+2(x-b)(x-d)$. This is quadratic with leading coefficient $1+2=3>0$.
    2. Evaluate at $x=a$: $f(a)=(a-a)(a-c)+2(a-b)(a-d)=2(a-b)(a-d)$. Since $a<b$ and $a<d$, both factors $(a-b),(a-d)$ are negative, so their product is positive: $f(a)>0$.
    3. Evaluate at $x=b$: $f(b)=(b-a)(b-c)+2(b-b)(b-d)=(b-a)(b-c)$. Here $b-a>0$ but $b-c<0$ (since $b<c$), so $f(b)<0$.
    4. Evaluate at $x=c$: $f(c)=(c-a)(c-c)+2(c-b)(c-d)=2(c-b)(c-d)$. Here $c-b>0$ but $c-d<0$ (since $c<d$), so $f(c)<0$.
    5. Evaluate at $x=d$: $f(d)=(d-a)(d-c)+2(d-b)(d-d)=(d-a)(d-c)$. Both factors positive (since $d$ is the largest), so $f(d)>0$. …

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