Q.Check whether the following sequence is a G.P. If so, write $t_n$: $\sqrt5, \dfrac{1}{\sqrt5}, \dfrac{1}{5\sqrt5}, \dfrac{1}{25\sqrt5}, \ldots$

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 $n=20$:

   $a_{20} = \dfrac52 \times \left(\dfrac12\right)^{19} = \dfrac{5}{2}\times\dfrac{1}{2^{19}} = \dfrac{5}{2^{20}}$

   ✓Final answer

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

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).

   $n$th term: $a_n = a\,r^{n-1} = \sqrt3\times(\sqrt3)^{n-1} = (\sqrt3)^n = 3^{n/2}$.

   Set this equal to 729. Since $729 = 3^6$:

   $3^{n/2} = 3^6 \implies \dfrac{n}{2} = 6 \implies n = 12$

   ✓Final answer

   (b) 12th.

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 $a_7 = a\,r^6 = 64$:

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

   Now compute $S_7 = a + ar + ar^2 + \cdots + ar^6$, i.e. the 7 terms $729, 486, 324, 216, 144, 96, 64$ (each obtained by multiplying the previous by $2/3$):

   $729+486=1215,\quad 1215+324=1539,\quad 1539+216=1755,$

   $1755+144=1899,\quad 1899+96=1995,\quad 1995+64=2059$

   So $S_7 = 2059$.

   ✓Final answer

   (d) 2059.

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:

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

   $x = \pm 2$

   Taking the standard positive value (consistent with all given terms being positive):

   $x = 2$

   (Check ratio: $\dfrac{2}{2/7} = 7$ and $\dfrac{14}{2}=7$ — consistent common ratio.)

   ✓Final answer

   $x = 2$ — option (b).

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.

   Here $a=2,\ b=8$, so $G = \sqrt{2\times 8} = \sqrt{16} = 4$.

   ✓Final answer

   The geometric mean of 2 and 8 is $4$, which is option (d).

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.

   (Checking $x=1$ also algebraically satisfies $x^2=ac$, with ratio $r=-\tfrac{7}{2}$, so both signs are valid roots of $x^2=ac$; $x=-1$ is the natural choice keeping the progression's terms uniformly signed.)

   ✓Final answer

   $x = \pm 1$ (with $x=-1$ giving the consistent common ratio $\tfrac{7}{2}$).

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

   $S_\infty=\frac{a}{1-r}=\frac{1}{1-\tfrac13}=\frac{1}{\tfrac23}=\frac32.$

   ✓Final answer

   Sum to infinity $=\dfrac{3}{2}.$

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

   Checking with the next pair: $a_3/a_2 = (1/81)/(-1/27) = -\dfrac{27}{81} = -\dfrac{1}{3}$, confirming a consistent ratio.

   ✓Final answer

   The correct option is (d) $-1/3$.

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 $r = 5/1 = 5$.

   The $n$-th term formula is $a_n = a\,r^{n-1}$. For $n=10$: $a_{10} = 1 \times 5^{10-1} = 5^9$.

   ✓Final answer

   The correct option is (a) $5^9$.

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 $\sqrt{ab}$.

    $\sqrt{44 \times 11} = \sqrt{484} = 22$.

    ✓Final answer

    The correct option is (d) 22.

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:

    $x^2 = \left(-\frac{2}{7}\right)\times\left(-\frac{7}{2}\right) = \frac{14}{14} = 1$

    $x = \pm 1$

    Both values are valid (a G.P. can have a negative common ratio), giving $x=1$ or $x=-1$.

    ✓Final answer

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

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 $r$.

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

    $\dfrac{a_6}{a_3} = \dfrac{ar^5}{ar^2} = r^3 = \dfrac{192}{24} = 8$

    So $r^3 = 8 \Rightarrow r = 2$.

    ✓Final answer

    $r = 2$ — option (b).