Q.Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true.

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\alpha, \beta, \gamma$ are the roots of the equation $3x^3 - 26x^2 + 52x - 24 = 0$ such that $\alpha, \beta, \gamma$ are in GP and $\alpha < \beta < \gamma$, then $3\alpha + 2\beta + \gamma =$ (A) $\dfrac{68}{3}$ (B) $\dfrac{56}{3}$ (C) $12$ (D) $24$

   ›Reveal solutionSolution

   Roots in GP means the middle root cubed equals the product of all three roots; that quickly pins down $\beta=2$, and factoring out $(x-2)$ gives the other two roots directly.

   Concept and Intuition

   If three numbers $a,\,ar,\,ar^2$ are in GP, their product is $a^3r^3=(ar)^3$ — exactly the cube of the middle term. So for a cubic whose roots are in GP, the middle root can be found immediately from the product-of-roots relation, without solving the full cubic first.

   Step-by-Step Solution

   1. For $3x^3-26x^2+52x-24=0$, product of roots $=-\dfrac{-24}{3}=8$.
   2. Since the roots are in GP, product $=\beta^3=8 \Rightarrow \beta=2$.
   3. Verify $x=2$ is a root: $3(8)-26(4)+52(2)-24=24-104+104-24=0$. ✓
   4. Divide the cubic by $(x-2)$ using synthetic division with coefficients $3,-26,52,-24$: quotient $3x^2-20x+12$.
   5. Solve $3x^2-20x+12=0$: $x=\dfrac{20\pm\sqrt{400-144}}{6}=\dfrac{20\pm16}{6}=6\ \text{or}\ \dfrac23$.
   6. So the three roots are $\dfrac23,\,2,\,6$; since $\alpha<\beta<\gamma$: $\alpha=\dfrac23,\ \beta=2,\ \gamma=6$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If the roots of the equation $x^3 - px^2 + qx - s = 0$ are in geometric progression, then (A) $q^2 = ps^2$ (B) $q^2 = p^2 s$ (C) $q^3 = ps^3$ (D) $q^3 = p^3 s$

   ›Reveal solutionSolution

   For a cubic with roots in geometric progression, the middle root is the geometric mean of the other two, and by relating the sum and product of roots we derive the condition $q^3 = p^3 s$. The correct option is (D).

   We are given the cubic $x^3 - p x^2 + q x - s = 0$ with roots in geometric progression (GP). That means if we denote the roots as $a/r,\ a,\ ar$ (where $a$ is the middle term and $r$ the common ratio), then the problem reduces to expressing the coefficients in terms of $a$ and $r$ and eliminating them.

   ---

   1. Set up the roots in GP form

   Let the three roots be

   $\frac{a}{r},\quad a,\quad a r.$

   This representation automatically ensures they are in GP: the ratio of consecutive terms is $r$.

   ---

   2. Use Vieta’s formulas

   For the cubic $x^3 - p x^2 + q x - s = 0$:

   • Sum of roots = $p$:

   $\frac{a}{r} + a + a r = p.$

   • Sum of pairwise products = $q$:

   $\frac{a}{r} \cdot a + a \cdot a r + \frac{a}{r} \cdot a r = q.$

   Simplify each term:

   $\frac{a^2}{r} + a^2 r + a^2 = q$.

   • Product of roots = $s$:

   $\frac{a}{r} \cdot a \cdot a r = a^3 = s.$

   ---

   3. Express everything in terms of $a$ and $r$

   From the product:

   $a^3 = s \quad\Rightarrow\quad a = s^{1/3}.$

   Now the sum of pairwise products:

   $\frac{a^2}{r} + a^2 r + a^2 = q.$

   Factor $a^2$:

   $a^2\left(\frac{1}{r} + r + 1\right) = q.$

   But $a^2 = (a^3)^{2/3} = s^{2/3}$. So

   $s^{2/3}\left(\frac{1}{r} + r + 1\right) = q. \qquad(1)$

   ---

   4. Eliminate $r$ using the sum of roots

   The sum of roots gives:

   $a\left(\frac{1}{r} + 1 + r\right) = p.$

   Since $a = s^{1/3}$, we have

   $s^{1/3}\left(\frac{1}{r} + r + 1\right) = p. \qquad(2)$

   ---

   5. Divide to eliminate the bracket

   Notice that both (1) and (2) contain the same factor $\left(\frac{1}{r} + r + 1\right)$. Divide (1) by (2):

   $$$$

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3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the sides a,b,c of the triangle ABC are in harmonic progression, then $\mathrm{cosec}^2\dfrac{A}{2}, \mathrm{cosec}^2\dfrac{B}{2}, \mathrm{cosec}^2\dfrac{C}{2}$ are in (A) Arithmetico-geometric progression (B) Arithmetic progression (C) Geometric progression (D) Harmonic progression

   ›Reveal solutionSolution

   $a,b,c$ in HP forces $\csc^2\frac{A}{2},\csc^2\frac{B}{2},\csc^2\frac{C}{2}$ to be in Arithmetic Progression.

   Concept and Intuition

   Half-angle cosecants connect directly to the triangle's sides through $\sin^2\frac{A}{2}=\dfrac{(s-b)(s-c)}{bc}$, so $\csc^2\frac{A}{2}=\dfrac{bc}{(s-b)(s-c)}$, and cyclically for $B,C$. "$a,b,c$ in HP" is really a statement about $\frac1a,\frac1b,\frac1c$ being in AP — the question is really asking what that reciprocal-AP condition forces onto these half-angle expressions.

   Step-by-Step Solution

   1. $a,b,c$ in HP $\Rightarrow \dfrac1a,\dfrac1b,\dfrac1c$ in AP $\Rightarrow \dfrac{2}{b}=\dfrac1a+\dfrac1c \Rightarrow 2ac=b(a+c)$.
   2. Write $\csc^2\frac{A}{2}=\dfrac{bc}{(s-b)(s-c)}$, $\csc^2\frac{B}{2}=\dfrac{ac}{(s-a)(s-c)}$, $\csc^2\frac{C}{2}=\dfrac{ab}{(s-a)(s-b)}$.
   3. Check with a concrete HP triple satisfying the triangle inequality, e.g. $\frac1a=2,\frac1b=3,\frac1c=4$ (so $a=\frac12,b=\frac13,c=\frac14$), which is a valid triangle.
   4. Computing $s=0.5417$, the three half-angle cosecant-squares come out to approximately $1.371,\;10.286,\;19.200$. …

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4. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If the roots of the equation $x^3 - 13x^2 + Kx - 27 = 0$ are in geometric progression then $K =$ (A) $-30$ (B) 30 (C) 39 (D) $-39$

   ›Reveal solutionSolution

   When roots are in GP, the middle term of the GP equals the cube root of the product of all three roots — plugging that known root into the cubic solves directly for the unknown coefficient $K$.

   Concept and Intuition

   For three numbers in geometric progression $\dfrac pr,\,p,\,pr$, their product is always $p^3$ regardless of the common ratio $r$ — the $r$'s cancel. So whenever a cubic's roots are stated to be in GP, the middle root is simply the cube root of the product of all roots (obtainable from Vieta's formula), letting us pin down one exact root without knowing the progression itself.

   Step-by-Step Solution

   1. For $x^3-13x^2+Kx-27=0$ (i.e. $a=-13,\,b=K,\,c=-27$ in $x^3+ax^2+bx+c$), the product of roots $\alpha\beta\gamma=-c=27$.
   2. Let roots be $\dfrac pr,p,pr$: product $=p^3=27\Rightarrow p=3$. …

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5. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If the roots of the equation $x^3 + ax^2 + bx + c = 0$ are in arithmetic progression, then (A) $a^3 - 3ab + c = 0$ (B) $9ab = 2a^3 + 27c$ (C) $a^2 - 2bc + c = 0$ (D) $3ab - 3c - a^3 = 0$

   ›Reveal solutionSolution

   The middle term of an AP is the average of all three roots, which for a cubic is always $-a/3$; plugging that back into the cubic gives the required relation. Answer: $9ab=2a^3+27c$.

   Concept and Intuition

   If three numbers are in AP, their middle term equals their arithmetic mean. For a cubic $x^3+ax^2+bx+c=0$, the sum of roots is always $-a$ (from Vieta's formulas), so if the roots are in AP the middle root must be $-a/3$ regardless of $b,c$ — and since it's an actual root, it must satisfy the cubic itself.

   Step-by-Step Solution

   1. Let roots be $p-d,\ p,\ p+d$. Sum $=3p=-a \Rightarrow p=-\dfrac{a}{3}$.
   2. Since $p$ is a root: $p^3+ap^2+bp+c=0$.
   3. Substitute $p=-a/3$: $-\dfrac{a^3}{27}+a\cdot\dfrac{a^2}{9}-\dfrac{ab}{3}+c=0 = -\dfrac{a^3}{27}+\dfrac{a^3}{9}-\dfrac{ab}{3}+c$. …

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6. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.The condition that the roots of $x^3 - bx^2 + cx - d = 0$ are in arithmetic progression is (A) $9cb = 2b^3 + 27d$ (B) $9cb = 2d^3 + 27b$ (C) $9cd = 2d^3 + 27b$ (D) $9cd = 2b^3 + 27d$

   ›Reveal solutionSolution

   For a cubic with roots in arithmetic progression, we set the roots as $p-q, p, p+q$, use Vieta’s formulas to relate sums and products to coefficients, and eliminate $p$ and $q$ to obtain the condition $9cd = 2b^3 + 27d$, which corresponds to option (D).

   We are given the cubic equation

   $x^3 - b x^2 + c x - d = 0$

   and told that its three roots are in arithmetic progression (AP). We need to find which relation among $b, c, d$ must hold.

   ---

   1. Represent the roots in AP

   If three numbers are in AP, we can denote them as

   $p - q,\quad p,\quad p + q$

   where $p$ is the middle term and $q$ is the common difference. This symmetric form simplifies algebra.

   ---

   2. Apply Vieta’s formulas

   For a cubic $x^3 - b x^2 + c x - d = 0$, Vieta gives:

   • Sum of roots: $(p-q) + p + (p+q) = 3p = b$ So

   $p = \frac{b}{3}$

   • Sum of pairwise products:

   $(p-q)p + p(p+q) + (p-q)(p+q) = p^2 - pq + p^2 + pq + p^2 - q^2 = 3p^2 - q^2 = c$

   • Product of roots:

   $(p-q) \cdot p \cdot (p+q) = p(p^2 - q^2) = d$

   ---

   3. Substitute $p = b/3$ into the other relations

   From the sum of pairwise products:

   $3\left(\frac{b}{3}\right)^2 - q^2 = c \quad\Rightarrow\quad \frac{b^2}{3} - q^2 = c$

   So

   $q^2 = \frac{b^2}{3} - c$

   From the product:

   $\frac{b}{3}\left( \left(\frac{b}{3}\right)^2 - q^2 \right) = d$

   Substitute $q^2$:

   $\frac{b}{3}\left( \frac{b^2}{9} - \left( \frac{b^2}{3} - c \right) \right) = d$

   Simplify inside parentheses:

   $\frac{b^2}{9} - \frac{b^2}{3} + c = \frac{b^2}{9} - \frac{3b^2}{9} + c = -\frac{2b^2}{9} + c$

   Thus:

   $\frac{b}{3} \left( c - \frac{2b^2}{9} \right) = d$

   ---

   4. Clear denominators and rearrange

   Multiply both sides by 3:

   $b \left( c - \frac{2b^2}{9} \right) = 3d$

   Multiply by 9:

   $b(9c - 2b^2) = 27d$

   So:

   $9bc - 2b^3 = 27d$

   Finally, bring terms together:

   $$$$

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7. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If 7 and 8 are the lengths of two sides of a triangle and 'a' is the length of its smallest side. The angles of the triangle are in AP and 'a' has two values $a_1$ and $a_2$ satisfying this condition. If $a_1 < a_2$ then $2a_1+3a_2 =$ (A) $15$ (B) $21$ (C) $24$ (D) $28$

   ›Reveal solutionSolution

   The angles of the triangle are in arithmetic progression, so they are $60^\circ - d,\ 60^\circ,\ 60^\circ + d$. Using the law of sines with sides 7, 8, and the smallest side $a$ yields two possible triangles, giving two values for $a$. Solving leads to $a_1 = 3$ and $a_2 = 5$, so $2a_1 + 3a_2 = 21$.

   ---

   Concept and Intuition

   When the angles of a triangle are in arithmetic progression (AP), the middle angle must be $60^\circ$ because the sum of angles is $180^\circ$. This is a classic fact: if three numbers are in AP, the middle one is the average. So here, the angles are $60^\circ - d$, $60^\circ$, $60^\circ + d$ for some $d \ge 0$.

   Now, the side opposite the smallest angle is the smallest side. Since $a$ is the smallest side, it must be opposite the smallest angle $60^\circ - d$. The other two sides are given as 7 and 8. Which side is opposite which angle? That depends on $d$. We use the law of sines to relate sides and sines of opposite angles. Because there are two possible assignments (7 opposite $60^\circ$ or 8 opposite $60^\circ$), we get two possible values for $a$.

   ---

   Step-by-step solution

   1. Set up the angles

      Let the angles be $A = 60^\circ - d$, $B = 60^\circ$, $C = 60^\circ + d$, with $d \ge 0$.

      Since $a$ is the smallest side, it is opposite the smallest angle $A = 60^\circ - d$.

   2. Apply the law of sines

   $\frac{a}{\sin(60^\circ - d)} = \frac{7}{\sin(\text{opposite angle of side 7})} = \frac{8}{\sin(\text{opposite angle of side 8})}$

   The two given sides 7 and 8 must be opposite the other two angles: $60^\circ$ and $60^\circ + d$.

   Since $7 < 8$, the side 7 is opposite the smaller of these two angles, i.e., $60^\circ$, and side 8 is opposite $60^\circ + d$.

   So:

   $\frac{a}{\sin(60^\circ - d)} = \frac{7}{\sin 60^\circ} = \frac{8}{\sin(60^\circ + d)}$

   3. Find $d$ from the equality of the two known ratios

   $\frac{7}{\sin 60^\circ} = \frac{8}{\sin(60^\circ + d)}$

   $\sin 60^\circ = \frac{\sqrt{3}}{2}$, so:

   $\frac{7}{\sqrt{3}/2} = \frac{8}{\sin(60^\circ + d)} \quad\Rightarrow\quad \frac{14}{\sqrt{3}} = \frac{8}{\sin(60^\circ + d)}$

   $\sin(60^\circ + d) = \frac{8\sqrt{3}}{14} = \frac{4\sqrt{3}}{7}$

   This gives two possible angles in $(0^\circ, 180^\circ)$:

   $60^\circ + d = \arcsin\!\left(\frac{4\sqrt{3}}{7}\right) \quad\text{or}\quad 60^\circ + d = 180^\circ - \arcsin\!\left(\frac{4\sqrt{3}}{7}\right)$

   Compute $\frac{4\sqrt{3}}{7} \approx 0.9897$, so $\arcsin(0.9897) \approx 81.79^\circ$.

   Hence:

   • Case 1: $60^\circ + d \approx 81.79^\circ \Rightarrow d \approx 21.79^\circ$
   • Case 2: $60^\circ + d \approx 180^\circ - 81.79^\circ = 98.21^\circ \Rightarrow d \approx 38.21^\circ$

   Both are valid because the angles remain positive and sum to $180^\circ$.

   4. Find $a$ for each case Using $\frac{a}{\sin(60^\circ - d)} = \frac{7}{\sin 60^\circ}$:

   $a = \frac{7 \sin(60^\circ - d)}{\sin 60^\circ} = \frac{7 \sin(60^\circ - d)}{\sqrt{3}/2} = \frac{14}{\sqrt{3}} \sin(60^\circ - d)$

   • Case 1: $d \approx 21.79^\circ$ $60^\circ - d \approx 38.21^\circ$, $\sin 38.21^\circ \approx 0.6186$ $a_1 \approx \frac{14}{\sqrt{3}} \times 0.6186 \approx \frac{14 \times 0.6186}{1.732} \approx 5.0$ So $a_1 = 5$. …

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8. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If the roots of the equation $4x^3 - 12x^2 + 11x + m = 0$ are in arithmetic progression, then $m =$ (A) $-3$ (B) $1$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   Roots in AP means they can be written $a-d,a,a+d$; matching Vieta's sum and pairwise-sum gives $a=1,d^2=1/4$, and the product-of-roots relation then gives $m=-3$.

   Concept and Intuition

   When a cubic's roots are said to be in arithmetic progression, the cleanest substitution is to write them symmetrically as $a-d,\ a,\ a+d$ — this makes the sum of roots collapse to just $3a$ (the $d$ cancels), instantly pinning down the middle root $a$ from the coefficients. The remaining Vieta relations (sum of pairwise products, product of roots) then determine $d$ and the unknown constant.

   Step-by-Step Solution

   1. Equation: $4x^3-12x^2+11x+m=0$. For $Ax^3+Bx^2+Cx+D=0$: sum of roots $=-B/A$, sum of pairwise products $=C/A$, product of roots $=-D/A$.
   2. Here $-B/A = 12/4=3$, $C/A=11/4$, $-D/A=-m/4$.
   3. Let roots be $a-d,\,a,\,a+d$. Sum $=3a=3 \Rightarrow a=1$. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.The values of x in $(-\pi,\pi)$ which satisfy the equation $8^{1+\cos^2 x+\cos^4 x+\cdots}=4^3$ are (A) $\pm\dfrac{\pi}{4},\ \pm\dfrac{3\pi}{4}$ (B) $\pm\dfrac{\pi}{6},\ \dfrac{\pi}{3}$ (C) $\pm\dfrac{\pi}{8}$ (D) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   This tests recognizing an infinite geometric series in the exponent and then solving the resulting equation for $\sin^2x$; the solutions are $\pm\pi/4,\pm3\pi/4$.

   Concept and Intuition

   The exponent $1+\cos^2x+\cos^4x+\cdots$ is a geometric series with first term $1$ and common ratio $\cos^2x$ (which lies in $[0,1)$ as long as $\sin x\ne0$), so it sums to $\dfrac{1}{1-\cos^2x}=\dfrac{1}{\sin^2x}$. This converts the transcendental-looking equation into a simple exponential equation once both sides are written with base 2.

   Step-by-Step Solution

   1. Sum the series: $1+\cos^2x+\cos^4x+\cdots = \dfrac{1}{1-\cos^2x}=\dfrac{1}{\sin^2x}$ (valid since $0\le\cos^2x<1$).
   2. Equation becomes $8^{1/\sin^2x}=4^3$. Write both sides base 2: $8=2^3$, $4^3=2^6$, so $2^{3/\sin^2x}=2^6$.
   3. Equate exponents: $\dfrac{3}{\sin^2x}=6 \Rightarrow \sin^2x=\dfrac12 \Rightarrow \sin x=\pm\dfrac{1}{\sqrt2}$. …

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10. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.The number of ways of selecting 3 numbers that are in GP from the set $\{1, 2, 3, \ldots, 100\}$ is (A) $18$ (B) $52$ (C) $14$ (D) $53$

    ›Reveal solutionSolution

    Count 3-term integer GPs $(a, ar, ar^2)$ with all terms $\le 100$, by summing over each possible integer common ratio $r=p\ge2$ the number of starting values $a=k$ with $kp^2\le100$.

    Concept and Intuition

    A 3-number GP within $\{1,\dots,100\}$ is determined by its smallest term $k$ and its common ratio $r$. If $r$ is a positive integer $p\ge2$, the triple is $(k, kp, kp^2)$, and we only need the largest term $kp^2$ to stay $\le 100$. So for each ratio $p$, the number of valid starting values $k$ is simply $\left\lfloor \dfrac{100}{p^2}\right\rfloor$ (since $k\ge1$ is otherwise unrestricted).

    Step-by-Step Solution

    1. Since the largest term is $kp^2 \le 100$ and $k\ge1$, we need $p^2\le100$, i.e. $p \le 10$. So $p$ ranges over $2,3,4,\dots,10$.
    2. For each $p$, the count of valid $k$ is $\left\lfloor\dfrac{100}{p^2}\right\rfloor$:
       • $p=2$: $\lfloor100/4\rfloor=25$
       • $p=3$: $\lfloor100/9\rfloor=11$
       • $p=4$: $\lfloor100/16\rfloor=6$
       • $p=5$: $\lfloor100/25\rfloor=4$
       • $p=6$: $\lfloor100/36\rfloor=2$
       • $p=7$: $\lfloor100/49\rfloor=2$ …

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11. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.In $\triangle ABC$, if a,b,c are in arithmetic progression and $C=2A$, then $a:c=$ (A) $4:5$ (B) $2:3$ (C) $5:6$ (D) $\sqrt3:2$

    ›Reveal solutionSolution

    The AP condition on the sides, combined with $C=2A$, forces $\cos A=\frac34$, giving $a:c=1:2\cos A=2:3$.

    Concept and Intuition

    Convert the side-length AP condition $2b=a+c$ into an angle condition using the Law of Sines, then substitute $C=2A$ (and hence $B=\pi-3A$) to get a single equation in $A$ alone. Solving that trig equation pins down $\cos A$, and the ratio $a:c$ follows immediately from $\sin A:\sin C=\sin A:\sin2A$.

    Step-by-Step Solution

    1. AP condition: $2b=a+c$. By Law of Sines ($a=k\sin A$, etc.): $2\sin B=\sin A+\sin C$.
    2. With $C=2A$: $B=\pi-A-C=\pi-3A$, so $\sin B=\sin3A$. The condition becomes $2\sin3A=\sin A+\sin2A$.
    3. Expand: $\sin3A=3\sin A-4\sin^3A$, $\sin2A=2\sin A\cos A$. So $2(3\sin A-4\sin^3A)=\sin A+2\sin A\cos A$.
    4. Divide by $\sin A$ (nonzero): $6-8\sin^2A=1+2\cos A$. Using $\sin^2A=1-\cos^2A$: $6-8+8\cos^2A=1+2\cos A\Rightarrow8\cos^2A-2\cos A-3=0$. …

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12. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.In $\triangle ABC$, if $(a-b)(s-c)=(b-c)(s-a)$, then $r_1,r_2,r_3$ are in (A) Arithmetic progression (B) Geometric progression (C) Harmonic progression (D) Arithmetico-geometric progression

    ›Reveal solutionSolution

    The given side/semi-perimeter condition is exactly the algebraic restatement of "$r_1,r_2,r_3$ are in AP". Answer: Arithmetic Progression.

    Concept and Intuition

    Rather than guessing which progression fits, translate the AP condition on the exradii ($2r_2=r_1+r_3$, equivalently $r_1-r_2=r_2-r_3$) into side/semi-perimeter language using $r_i=\Delta/(s-\text{side}_i)$, and check whether it matches the given condition verbatim.

    Step-by-Step Solution

    1. $r_1-r_2=\dfrac{\Delta}{s-a}-\dfrac{\Delta}{s-b}=\dfrac{\Delta[(s-b)-(s-a)]}{(s-a)(s-b)}=\dfrac{\Delta(a-b)}{(s-a)(s-b)}$.
    2. Similarly, $r_2-r_3=\dfrac{\Delta(b-c)}{(s-b)(s-c)}$.
    3. The AP condition $r_1-r_2=r_2-r_3$ becomes $\dfrac{\Delta(a-b)}{(s-a)(s-b)}=\dfrac{\Delta(b-c)}{(s-b)(s-c)}$.
    4. Cancel $\Delta$ and $(s-b)$ from both sides: $\dfrac{a-b}{s-a}=\dfrac{b-c}{s-c}$, i.e. $(a-b)(s-c)=(b-c)(s-a)$. …

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