Q.Find three numbers in G.P. such that their sum is 21 and sum of their squares is 189.

Showing the 12 most recent of 40 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.20th term of the G.P. $\dfrac{5}{2}, \dfrac{5}{4}, \dfrac{5}{8}, \ldots$

   (a) $\dfrac{5}{2^{20}}$

   (b) $\dfrac{1}{2^{20}}$

   (c) $\dfrac{5}{2^{19}}$

   (d) $\dfrac{5}{2^{18}}$

   ›Reveal solutionSolution

   Identify the first term and common ratio, then apply $a_n = a\,r^{n-1}$ for $n=20$.

   The G.P. is $\dfrac52, \dfrac54, \dfrac58, \ldots$ with first term $a = \dfrac52$ and common ratio $r = \dfrac{5/4}{5/2} = \dfrac12$.

   The $n$th term of a G.P. is $a_n = a\,r^{n-1}$. For $n=20$: …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which term of the sequence $\sqrt{3}, 3, 3\sqrt{3}, \ldots$ is 729?

   (a) 9th

   (b) 12th

   (c) 13th

   (d) 11th

   ›Reveal solutionSolution

   The sequence is a G.P. with common ratio $\sqrt3$; express the $n$th term as a power of $\sqrt3$ and match it to $729$.

   The sequence $\sqrt3, 3, 3\sqrt3, \ldots$ has first term $a=\sqrt3$ and common ratio $r = \dfrac{3}{\sqrt3} = \sqrt3$ (check: $3\sqrt3/3=\sqrt3$ too).

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Given a G.P. with $a = 729$ and 7th term 64, then $S_7 = ?$

   (a) 2187

   (b) 64

   (c) 2159

   (d) 2059

   ›Reveal solutionSolution

   First find the common ratio from $a\,r^6 = 64$, then sum the first 7 terms directly.

   Given $a = 729$ and the 7th term $a_7 = a\,r^6 = 64$:

   $729\,r^6 = 64 \implies r^6 = \dfrac{64}{729} = \left(\dfrac23\right)^6 \implies r = \dfrac23$

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The value of x for which the progression 2/7, x, 14 is in G.P., is:

   (a) 4

   (b) 2

   (c) 1

   (d) None of these

   ›Reveal solutionSolution

   In a G.P., each term squared (the middle term) equals the product of its neighbours; solving gives x = 2.

   For $\dfrac{2}{7}, x, 14$ to be in G.P., the middle term squared must equal the product of the outer terms:

   $x^2 = \frac{2}{7} \times 14 = 4$

   $x = \pm 2$

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.The Geometric mean of 2 and 8 is:

   (a) $-4$

   (b) $5$

   (c) $-5$

   (d) $4$

   ›Reveal solutionSolution

   The geometric mean of two numbers $a,b$ is $\sqrt{ab}$; for 2 and 8 this gives $\sqrt{16}=4$.

   The geometric mean (GM) of two positive numbers $a$ and $b$ is defined as $G=\sqrt{ab}$, the number such that $a, G, b$ form a G.P.

   …

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6. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: If $-\dfrac{2}{7}, x, -\dfrac{7}{2}$ are consecutive terms of a geometric progression, then the value of $x$ will be ______.

   ›Reveal solutionSolution

   If $a, x, c$ are consecutive terms of a G.P., then $x^2 = a\cdot c$; here that gives $x=\pm1$, and $x=-1$ is the value consistent with a single common ratio across the progression.

   For three consecutive G.P. terms $a, x, c$: $\dfrac{x}{a} = \dfrac{c}{x} \Rightarrow x^2 = ac$.

   Here $a = -\dfrac{2}{7}$, $c = -\dfrac{7}{2}$, so $x^2 = \left(-\dfrac{2}{7}\right)\left(-\dfrac{7}{2}\right) = 1$, giving $x = \pm 1$.

   Checking $x=-1$: common ratio $r = \dfrac{x}{a} = \dfrac{-1}{-2/7} = \dfrac{7}{2}$, and indeed $x \cdot r = -1 \times \dfrac{7}{2} = -\dfrac{7}{2} = c$ ✓ — a single positive ratio $\tfrac{7}{2}$ carries all three (negative) terms consistently.

   …

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7. CBSE 2026Set 1A1 mark
   Q.Find the sum to infinity in Geometric Progression $1, \frac{1}{3}, \frac{1}{9}, \ldots$

   ›Reveal solutionSolution

   With $a=1$, $r=\tfrac13$, $S_\infty=\dfrac{1}{1-\frac13}=\dfrac32$.

   The GP $1,\tfrac13,\tfrac19,\dots$ has first term $a=1$ and common ratio $r=\tfrac13$ (with $|r|<1$). The sum to infinity is …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.The common ratio of the geometrical progression $\dfrac{1}{9}, \dfrac{-1}{27}, \dfrac{1}{81}, \dfrac{-1}{243}, \ldots$ is

   (a) $1/3$

   (b) 3

   (c) -3

   (d) $-1/3$

   ›Reveal solutionSolution

   The common ratio $r = -\dfrac{1}{3}$.

   For a G.P., $r = \dfrac{a_2}{a_1}$.

   Here $a_1 = \dfrac{1}{9}$, $a_2 = -\dfrac{1}{27}$: $r = \dfrac{-1/27}{1/9} = -\dfrac{1}{27}\times\dfrac{9}{1} = -\dfrac{9}{27} = -\dfrac{1}{3}$.

   …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.The 10th term of the geometrical progression $1, 5, 25, 125, \ldots$ is

   (a) $5^9$

   (b) $5^{10}$

   (c) $5^{11}$

   (d) $5^{12}$

   ›Reveal solutionSolution

   The 10th term is $5^9$.

   For the G.P. $1, 5, 25, 125, \ldots$: first term $a=1$, common ratio $r = 5/1 = 5$.

   …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The geometrical mean of 44 and 11 is

    (a) 27.5

    (b) 25

    (c) 55

    (d) 22

    ›Reveal solutionSolution

    The geometric mean of 44 and 11 is $22$.

    For two positive numbers $a$ and $b$, the geometric mean is $\sqrt{ab}$.

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.For what value of $x$ are the numbers $\dfrac{-2}{7}, x, \dfrac{-7}{2}$ in G.P.?

    (a) $x = 1, x = -1$

    (b) $x = -1, x = 3$

    (c) $x = 0, x = 2$

    (d) None of these

    ›Reveal solutionSolution

    For three numbers in G.P., the middle term squared equals the product of the outer terms: $x^2 = \left(-\frac27\right)\left(-\frac72\right) = 1$, so $x=\pm1$.

    If $-\dfrac27, x, -\dfrac72$ are in G.P., then the middle term is the geometric mean of the other two: …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.If third term of a G.P. is 24 and 6th term is 192, then its common ratio is:

    (a) 1/2

    (b) 2

    (c) 6

    (d) None of these

    ›Reveal solutionSolution

    Dividing the 6th term by the 3rd term of a G.P. gives $r^3$.

    Let the G.P. have first term $a$ and common ratio $r$.

    $a_3 = ar^2 = 24$, $a_6 = ar^5 = 192$.

    …

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