Q.Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.Every term of a geometric progression is positive, and every term is the sum of the two preceding terms. Then the common ratio of the geometric progression is: (A) $\frac{1+\sqrt{5}}{2}$ (B) $\frac{\sqrt{5}-1}{2}$ (C) 1 (D) $\frac{1-\sqrt{5}}{2}$

   ›Reveal solutionSolution

   The problem gives a geometric progression where each term equals the sum of the two preceding terms. This leads to a quadratic equation for the common ratio $r$, and since all terms are positive, the only valid solution is $r = \frac{1+\sqrt{5}}{2}$, which is option (A).

   We have a geometric progression (GP) where every term is positive. Let the first term be $a > 0$ and the common ratio be $r > 0$. Then the terms are:

   $a,\; ar,\; ar^2,\; ar^3,\; \dots$

   The condition says: every term is the sum of the two preceding terms. That means for any term starting from the third one, we have:

   $ar^2 = a + ar$

   (Since the third term equals the sum of the first two terms. If this holds for the third term, it will automatically hold for all later terms because multiplying through by powers of $r$ preserves the relation.)

   1. Set up the equation From $ar^2 = a + ar$, divide through by $a$ (which is positive, so safe):

   $r^2 = 1 + r$

   2. Rearrange into standard quadratic form

   $r^2 - r - 1 = 0$

   3. Solve the quadratic Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1,\; b=-1,\; c=-1$:

   $r = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

   4. Apply the positivity condition

      The two possible values are:

      • $\frac{1 + \sqrt{5}}{2} \approx 1.618$ (positive)
      • $\frac{1 - \sqrt{5}}{2} \approx -0.618$ (negative) …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.The product of three numbers in geometric progression is 8 and the sum of the product of the numbers taken in pairs is 14 . Find the numbers. (A) $4,2,1 \text { and } 1,2,4$ (B) $\frac{1}{4}, \frac{1}{2}, 1$ and $1, \frac{1}{2}, \frac{1}{4}$ (C) $4 \sqrt{2}, \sqrt{2}, 1$ and $1, \sqrt{2}, 4 \sqrt{2}$ (D) $-4,-2,1$ and $1,-2,-4$

   ›Reveal solutionSolution

   For three numbers in geometric progression, set them as $\frac{a}{r}, a, ar$. Their product gives $a^3 = 8 \Rightarrow a=2$. The sum of pairwise products yields a quadratic in $r$, giving two symmetric sequences: $4,2,1$ and $1,2,4$. The correct option is (A).

   Concept & Intuition

   When a problem says "three numbers in geometric progression," the cleverest move is not to use $a, ar, ar^2$ — that leads to messy algebra. Instead, use $\frac{a}{r}, a, ar$. Why? Because the product becomes $a^3$ instantly (the $r$'s cancel), and the sum of pairwise products becomes symmetric and easy to handle. This trick turns a system of equations into a single quadratic.

   Step-by-step solution

   1. Set up the numbers Let the three numbers be $\frac{a}{r}$, $a$, and $ar$, where $a$ is the middle term and $r$ is the common ratio. Their product is:

   $\frac{a}{r} \cdot a \cdot ar = a^3$

   Given that the product is $8$, we have:

   $a^3 = 8 \quad \Rightarrow \quad a = 2$

   (We take the real cube root; $a=2$ is the only real possibility.)

   2. Sum of pairwise products The three pairwise products are:

   $\frac{a}{r} \cdot a = \frac{a^2}{r}, \quad \frac{a}{r} \cdot ar = a^2, \quad a \cdot ar = a^2 r$

   Their sum is:

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

   Substitute $a=2$ (so $a^2 = 4$):

   $4 \left( r + 1 + \frac{1}{r} \right) = 14$

   Divide both sides by $2$ to simplify:

   $2 \left( r + 1 + \frac{1}{r} \right) = 7$

   So:

   $r + \frac{1}{r} + 1 = \frac{7}{2} \quad \Rightarrow \quad r + \frac{1}{r} = \frac{5}{2}$

   3. Solve for $r$ Multiply through by $r$:

   $r^2 + 1 = \frac{5}{2} r \quad \Rightarrow \quad 2r^2 - 5r + 2 = 0$

   Factor:

   $(2r - 1)(r - 2) = 0$

   So $r = 2$ or $r = \frac{1}{2}$.

   4. Find the two sequences …

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3. KCET 2026Set UNKNOWN1 markMCQ
   Q.If we insert two numbers between $\sqrt{2}$ and $4$ so that the resulting sequence is in G.P., then the inserted numbers in the order are (A) $4, \sqrt{2}$ (B) $2, 2\sqrt{2}$ (C) $\sqrt{8}, 2$ (D) $2\sqrt{2}, 4$

   ›Reveal solutionSolution

   With $\sqrt{2}, a, b, 4$ forming a 4-term G.P., find the common ratio $r$ from $4 = \sqrt{2}\cdot r^3$, then compute $a=\sqrt{2}\,r$ and $b=\sqrt{2}\,r^2$.

   Step 1 — Set up the G.P. with common ratio $r$

   Let the four terms in G.P. be $\sqrt{2},\ a,\ b,\ 4$, so

   $a = \sqrt{2}\,r, \qquad b = \sqrt{2}\,r^2, \qquad 4 = \sqrt{2}\,r^3.$

   Step 2 — Solve for the common ratio

   $r^3 = \frac{4}{\sqrt{2}} = \frac{2^2}{2^{1/2}} = 2^{3/2}.$

   Taking the cube root:

   $r = 2^{1/2} = \sqrt{2}.$

   Step 3 — Compute the two inserted numbers

   $a = \sqrt{2}\cdot \sqrt{2} = 2,$ …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.The terms of an infinitely decreasing geometric progression in which all the terms are positive, the first term is $\mathbf{4}$, and the difference between third and fifth term is $\frac{32}{81}$, then which of the following is not true (A) $S_{\infty}=3+2 \sqrt{2}$ (B) $r=\frac{1}{3}$ (C) $S_{\infty}=6$ (D) $r=\frac{2 \sqrt{2}}{3}$

   ›Reveal solutionSolution

   Solving $4r^2-4r^4=\tfrac{32}{81}$ gives $r=\tfrac13$ (with $S_\infty=6$) or $r=\tfrac{2\sqrt2}{3}$ (with $S_\infty=36+24\sqrt2$). The value $S_\infty=3+2\sqrt2$ fits neither, so the statement that is not true is option (A).

   For a decreasing GP with all positive terms, $a=4$ and $0<r<1$. The $n$-th term is $ar^{n-1}$.

   Given condition ($T_3-T_5=\tfrac{32}{81}$):

   $4r^2-4r^4=\frac{32}{81}\;\Rightarrow\;r^2-r^4=\frac{8}{81}$

   Solve the quadratic in $x=r^2$:

   $x^2-x+\frac{8}{81}=0\;\Rightarrow\;x=\frac{1\pm\sqrt{1-\tfrac{32}{81}}}{2}=\frac{1\pm\sqrt{49/81}}{2}=\frac{1\pm\tfrac79}{2}$

   $x=\frac{8}{9}\quad\text{or}\quad x=\frac{1}{9}\;\Rightarrow\; r=\frac{2\sqrt2}{3}\ \text{or}\ r=\frac13$

   Both lie in $(0,1)$, so both give valid decreasing GPs. Evaluate each option:

   • (B) $r=\tfrac13$ — true (one valid root). …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A geometric progression consists of an even number of terms. If the sum of all the terms is five times the sum of the terms occupying the odd places, then the common ratio of the geometric progression is (A) $r=4$ (B) $r=3$ (C) $r=6$ (D) $r=2$

   ›Reveal solutionSolution

   The key idea is to separate the GP into its odd-position terms (which themselves form a GP) and even-position terms, then use the given sum condition to solve for the common ratio. The answer is $r = 4$.

   ---

   We are told the GP has an even number of terms. Let’s say there are $2n$ terms. The terms are:

   $a,\; ar,\; ar^2,\; ar^3,\; \dots,\; ar^{2n-1}$

   The “odd places” mean positions 1, 3, 5, …, i.e., the first, third, fifth, … terms. These are:

   $a,\; ar^2,\; ar^4,\; \dots,\; ar^{2n-2}$

   Notice that these odd-position terms themselves form a GP with first term $a$ and common ratio $r^2$, and there are $n$ of them.

   Similarly, the even-position terms (positions 2, 4, 6, …) are:

   $ar,\; ar^3,\; ar^5,\; \dots,\; ar^{2n-1}$

   which is also a GP with first term $ar$ and common ratio $r^2$, also $n$ terms.

   ---

   Step-by-step reasoning:

   1. Sum of all terms The sum of the full GP (first term $a$, ratio $r$, $2n$ terms) is:

   $S_{\text{all}} = a \cdot \frac{r^{2n} - 1}{r - 1}$

   2. Sum of odd-place terms The odd-place GP has first term $a$, ratio $r^2$, and $n$ terms:

   $S_{\text{odd}} = a \cdot \frac{(r^2)^n - 1}{r^2 - 1} = a \cdot \frac{r^{2n} - 1}{r^2 - 1}$

   3. Given condition The problem says:

   $S_{\text{all}} = 5 \cdot S_{\text{odd}}$

   Substituting:

   $a \cdot \frac{r^{2n} - 1}{r - 1} = 5 \cdot a \cdot \frac{r^{2n} - 1}{r^2 - 1}$

   4. Cancel common factors …

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6. COMEDK 2024Set 2024-A1 markMCQ
   Q.The sum of four numbers in a geometric progression is 60 , and the arithmetic mean of the first and the last number is 18 . Then the numbers are (A) $10,8,16,26$ (B) $32,16,4,8$ (C) $32,16,8,2$ (D) $4,8,16,32$

   ›Reveal solutionSolution

   We set up a geometric progression $a, ar, ar^2, ar^3$ with sum 60 and $\frac{a+ar^3}{2}=18$. Solving gives $a=4, r=2$, so the numbers are $4,8,16,32$ — option (D).

   We have four numbers in geometric progression. Let them be

   $a,\; ar,\; ar^2,\; ar^3$

   where $a$ is the first term and $r$ the common ratio.

   The problem gives two conditions:

   1. Sum of the four numbers is 60:

   $a + ar + ar^2 + ar^3 = 60$

   2. Arithmetic mean of the first and last is 18:

   $\frac{a + ar^3}{2} = 18 \quad\Rightarrow\quad a + ar^3 = 36$

   ---

   Step-by-step reasoning

   1. Use the second condition to simplify From $a + ar^3 = 36$, we have

   $a(1 + r^3) = 36 \qquad(1)$

   2. Rewrite the sum using the second condition The sum is

   $a + ar + ar^2 + ar^3 = (a + ar^3) + (ar + ar^2) = 36 + ar(1 + r)$

   Since the total sum is 60, we get

   $36 + ar(1 + r) = 60 \quad\Rightarrow\quad ar(1 + r) = 24 \qquad(2)$

   3. Divide equation (2) by equation (1) to eliminate $a$

   $\frac{ar(1 + r)}{a(1 + r^3)} = \frac{24}{36}$

   Simplifying:

   $\frac{r(1 + r)}{1 + r^3} = \frac{2}{3}$

   Recall the identity $1 + r^3 = (1 + r)(1 - r + r^2)$. Cancel $(1+r)$ (assuming $r \neq -1$, which is fine here):

   $\frac{r}{1 - r + r^2} = \frac{2}{3}$

   4. Cross-multiply and solve for $r$

   $3r = 2(1 - r + r^2) \quad\Rightarrow\quad 3r = 2 - 2r + 2r^2$

   $0 = 2 - 5r + 2r^2 \quad\Rightarrow\quad 2r^2 - 5r + 2 = 0$

   Factor:

   $(2r - 1)(r - 2) = 0$

   So $r = \frac{1}{2}$ or $r = 2$.

   5. Find $a$ for each case …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\text { If } 6^{\text {th }} \text { term of a geometric progression is }-\frac{1}{32} \text { and } 9^{\text {th }} \text { term is } \frac{1}{256} \text { then } r \text { is }$

   (A) 2 (B) $-\frac{1}{2}$ (C) $\frac{1}{2}$ (D) $-2$

   ›Reveal solutionSolution

   The common ratio $r$ is found by dividing the 9th term by the 6th term, giving $r^3 = -\frac{1}{8}$, so $r = -\frac{1}{2}$. The correct option is (B).

   Concept & Intuition

   In a geometric progression (GP), each term is the previous term multiplied by the common ratio $r$. So if you know two terms that are a fixed number of steps apart, you can find $r$ by taking the appropriate root of their ratio. Here, the 9th term is three steps after the 6th term, so the ratio of the 9th to the 6th term equals $r^3$. Once we compute that ratio, we solve for $r$, being careful with signs because the terms have opposite signs — that tells us $r$ must be negative.

   Step-by-step solution

   1. Write the general term formula For a GP with first term $a$ and common ratio $r$, the $n$-th term is

   $T_n = a r^{n-1}.$

   So the 6th term is $T_6 = a r^5$ and the 9th term is $T_9 = a r^8$.

   2. Set up the given values We know

   $a r^5 = -\frac{1}{32} \quad \text{and} \quad a r^8 = \frac{1}{256}.$

   3. Divide the two equations to eliminate $a$

   $\frac{T_9}{T_6} = \frac{a r^8}{a r^5} = r^3.$

   Substituting the numbers:

   $r^3 = \frac{\frac{1}{256}}{-\frac{1}{32}} = \frac{1}{256} \times \left(-\frac{32}{1}\right) = -\frac{32}{256} = -\frac{1}{8}.$

   4. Solve for $r$ Take the cube root of both sides: $$$$

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.Consider an infinite geometric series with first term '$a$' and common ratio '$r$'. If the sum of infinite geometric series is 4 and the second term is $\frac{3}{4}$ then (A) $a=1 \quad r=-\frac{3}{4}$ (B) $a=3 \quad r=\frac{1}{4}$ (C) $a=-3 \quad r=-\frac{1}{4}$ (D) $a=-1 \quad r=\frac{3}{4}$

   ›Reveal solutionSolution

   The problem gives two conditions: the infinite sum is 4 and the second term is 3/4. Using the formulas for sum to infinity and the nth term, we solve for a and r, then check which option matches.

   We have an infinite geometric series with first term $a$ and common ratio $r$. The sum to infinity exists only if $|r| < 1$, and is given by

   $S_\infty = \frac{a}{1 - r}$

   We are told this sum is 4, so

   $\frac{a}{1 - r} = 4 \quad \Rightarrow \quad a = 4(1 - r) \qquad(1)$

   The second term of a geometric series is $ar$. We are told it equals $\frac{3}{4}$, so

   $ar = \frac{3}{4} \qquad(2)$

   Now we solve these two equations together.

   1. Substitute $a$ from (1) into (2):

   $4(1 - r) \cdot r = \frac{3}{4}$

   Simplify:

   $4r(1 - r) = \frac{3}{4}$

   Multiply both sides by 4:

   $16r(1 - r) = 3$

   Expand:

   $16r - 16r^2 = 3$

   Bring all terms to one side:

   $-16r^2 + 16r - 3 = 0$

   Multiply by -1:

   $16r^2 - 16r + 3 = 0$

   2. Solve this quadratic for $r$. Use the quadratic formula:

   $r = \frac{16 \pm \sqrt{256 - 192}}{32} = \frac{16 \pm \sqrt{64}}{32} = \frac{16 \pm 8}{32}$

   So the two possibilities are:

   $r = \frac{24}{32} = \frac{3}{4} \quad \text{or} \quad r = \frac{8}{32} = \frac{1}{4}$

   3. For each $r$, find $a$ using (1):

   • If $r = \frac{3}{4}$, then $a = 4\left(1 - \frac{3}{4}\right) = 4 \cdot \frac{1}{4} = 1$.
   • If $r = \frac{1}{4}$, then $a = 4\left(1 - \frac{1}{4}\right) = 4 \cdot \frac{3}{4} = 3$.

   4. Check the convergence condition: For an infinite sum to exist, we need $|r| < 1$. Both $\frac{3}{4}$ and $\frac{1}{4}$ satisfy this, so both pairs $(a=1, r=\frac{3}{4})$ and $(a=3, r=\frac{1}{4})$ are mathematically valid.

   Now look at the options: …

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9. COMEDK 2024Set 2024-E1 markMCQ
   Q.A number consists of three digits in geometric progression. The sum of the right hand and left hand digits exceeds twice the middle digit by 1 and the sum of left hand and middle digits is two third of the sum of the middle and right hand digits. Then the sum of digits of number is (A) $\frac{1}{4}$ (B) 19 (C) 469 (D) 109

   ›Reveal solutionSolution

   The problem gives a three-digit number with digits in geometric progression and two conditions linking the digits. Solving the system yields the digits 4, 6, 9, so the number is 469 and the sum of its digits is 19. The correct option is (B).

   We have a three-digit number. Let its hundreds digit be $a$, tens digit be $b$, and units digit be $c$. The digits are in geometric progression, so $b^2 = a c$. Also, the number is $100a + 10b + c$.

   Concept and intuition:

   When digits are in geometric progression, we can write them as $a$, $ar$, $ar^2$ for some common ratio $r$. The two given conditions then become equations in $a$ and $r$. Solving them gives the digits, and then we sum them. The trick is to handle the "exceeds by 1" and "two thirds" statements carefully.

   ---

   1. Set up the geometric progression Let the digits be $a$, $ar$, $ar^2$ where $a$ is a positive integer from 1 to 9 (hundreds digit cannot be 0) and $r$ is a positive rational number (since digits are integers, $r$ must be rational). So:

   $b = ar, \quad c = ar^2.$

   2. Translate the first condition "The sum of the right hand and left hand digits exceeds twice the middle digit by 1" means:

   $(c + a) = 2b + 1.$

   Substitute $b = ar$, $c = ar^2$:

   $ar^2 + a = 2ar + 1.$

   Divide through by $a$ (since $a > 0$):

   $r^2 + 1 = 2r + \frac{1}{a}. \qquad(1)$

   3. Translate the second condition "The sum of left hand and middle digits is two thirds of the sum of the middle and right hand digits" means:

   $a + b = \frac{2}{3}(b + c).$

   Substitute:

   $a + ar = \frac{2}{3}(ar + ar^2).$

   Factor $a$ (nonzero):

   $1 + r = \frac{2}{3}(r + r^2).$

   Multiply both sides by 3:

   $3 + 3r = 2r + 2r^2.$

   Rearranging:

   $0 = 2r^2 - r - 3.$

   Solve:

   $2r^2 - r - 3 = 0 \quad \Rightarrow \quad (2r - 3)(r + 1) = 0.$

   So $r = \frac{3}{2}$ or $r = -1$. Since digits are positive, $r = \frac{3}{2}$.

   4. Find $a$ using equation (1) Plug $r = \frac{3}{2}$ into (1): $$$$

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10. COMEDK 2024Set 2024-M1 markMCQ
    Q.A geometric progression consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying odd places, then the common ratio of the G.P is (A) 3 (B) 2 (C) 5 (D) 4

    ›Reveal solutionSolution

    The key idea is to treat the odd‑placed terms as a separate geometric progression and use the given ratio of sums to solve for the common ratio. The common ratio is found to be 4.

    We are told the GP has an even number of terms. Let the number of terms be $2n$.

    The terms are:

    $a,\; ar,\; ar^2,\; ar^3,\; \dots,\; ar^{2n-1}$

    The sum of all terms is

    $S_{\text{all}} = a\frac{r^{2n}-1}{r-1}$

    The odd‑placed terms are the 1st, 3rd, 5th, … terms:

    $a,\; ar^2,\; ar^4,\; \dots,\; ar^{2n-2}$

    These form a GP with first term $a$ and common ratio $r^2$, and there are $n$ of them. Their sum is

    $S_{\text{odd}} = a\frac{(r^2)^n - 1}{r^2 - 1} = a\frac{r^{2n} - 1}{r^2 - 1}$

    The problem states:

    $S_{\text{all}} = 5 \cdot S_{\text{odd}}$

    Substitute:

    $a\frac{r^{2n}-1}{r-1} = 5 \cdot a\frac{r^{2n}-1}{r^2-1}$

    Since $a \neq 0$ and $r^{2n} \neq 1$ (otherwise the sums would be zero or trivial), we cancel $a(r^{2n}-1)$:

    $\frac{1}{r-1} = \frac{5}{r^2-1}$

    Recall $r^2 - 1 = (r-1)(r+1)$. So:

    $\frac{1}{r-1} = \frac{5}{(r-1)(r+1)}$

    Multiply both sides by $(r-1)$ (which is nonzero, else the GP would be constant and the condition fails):

    $1 = \frac{5}{r+1}$

    Thus:

    $$$$

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11. COMEDK 2024Set 2024-M1 markMCQ
    Q.$$$$

    (32) \times(32)^{\frac{1}{6}} \times(32)^{\frac{1}{36}} \times-----\infty \text { is equal to } $$ (A) 16 (B) 32 (C) 0 (D) 64

    ›Reveal solutionSolution

    The infinite product of powers of 32 can be rewritten as a single power of 32 whose exponent is an infinite geometric series. Summing the series gives exponent 6/5, so the result is $32^{6/5} = (2^5)^{6/5} = 2^6 = 64$. The correct option is (D).

    The key insight is that multiplying powers of the same base means adding exponents. Here we have an infinite product:

    $32 \times 32^{1/6} \times 32^{1/36} \times 32^{1/216} \times \cdots$

    The exponents form a geometric sequence: $1, \frac{1}{6}, \frac{1}{36}, \frac{1}{216}, \dots$ with common ratio $r = \frac{1}{6}$. Since $|r| < 1$, the infinite sum of exponents converges. The whole expression becomes $32^{\text{(sum of exponents)}}$.

    1. Write the product as a single power

    $32^1 \times 32^{1/6} \times 32^{1/36} \times \cdots = 32^{\,1 + \frac{1}{6} + \frac{1}{36} + \frac{1}{216} + \cdots}$

    2. Sum the infinite geometric series The series is $S = 1 + \frac{1}{6} + \frac{1}{36} + \frac{1}{216} + \cdots$ First term $a = 1$, common ratio $r = \frac{1}{6}$. For an infinite geometric series with $|r| < 1$:

    $S = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{6}} = \frac{1}{\frac{5}{6}} = \frac{6}{5}$

    3. Evaluate the power $$$$

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12. KCET 2019Set A-11 markMCQ
    Q.The third term of a G.P. is 9. The product of its first five terms is (A) $3^5$ (B) $3^9$ (C) $3^{10}$ (D) $3^{12}$

    ›Reveal solutionSolution

    In a geometric progression, the product of the first five terms equals the fifth power of the middle (third) term. Since the third term is 9, the product is $9^5 = 3^{10}$.

    The key insight here is symmetry. In any geometric progression with an odd number of terms, the middle term acts as a geometric mean of the entire set. For five terms, the third term is exactly the geometric mean of the first and fifth, the second and fourth, and so on. This means the product of all five terms simplifies beautifully to the third term raised to the power of 5.

    Let’s see why this works.

    1. Set up the general G.P. Let the first term be $a$ and the common ratio be $r$. Then the first five terms are:

    $a,\ ar,\ ar^2,\ ar^3,\ ar^4$

    2. Use the given information.

       The third term is $ar^2 = 9$.

    3. Write the product of the first five terms.

    $P = a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4$

    Count the powers of $a$: there are five $a$'s, so $a^5$.

    Count the powers of $r$: the exponents are $0+1+2+3+4 = 10$, so $r^{10}$.

    Thus

    $P = a^5 r^{10}$

    4. Rewrite in terms of the third term. Notice that $a^5 r^{10} = (a r^2)^5$. Why? Because $(a r^2)^5 = a^5 (r^2)^5 = a^5 r^{10}$. And $a r^2$ is exactly the third term, which is 9. So …

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