Q.The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

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).