Q.Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

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)

      Since every term of the GP is positive, the common ratio must be positive. Therefore, we discard the negative root.

   Watch out

   A common mistake is to forget that the common ratio of a GP with all positive terms must be positive. The negative root $\frac{1-\sqrt{5}}{2}$ is mathematically valid from the equation but violates the positivity condition.

   Tip

   The positive root $\frac{1+\sqrt{5}}{2}$ is the famous golden ratio $\phi$. This problem shows a beautiful link between geometric progressions and the golden ratio.

   Thus, the common ratio is $\frac{1+\sqrt{5}}{2}$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

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

      • If $r = 2$: numbers are $\frac{2}{2}, 2, 2\cdot 2 = 1, 2, 4$.
      • If $r = \frac{1}{2}$: numbers are $\frac{2}{1/2}, 2, 2\cdot\frac{1}{2} = 4, 2, 1$. These are just reversals of each other.

   5. Match with options

      Option (A) lists exactly $4,2,1$ and $1,2,4$.

      Option (B) has fractions, (C) has $\sqrt{2}$, (D) has negatives — none satisfy the product $8$ and sum $14$ simultaneously.

   Watch out

   A common mistake is to use $a, ar, ar^2$ and then solve a cubic in $r$ — it works but is unnecessarily messy. The symmetric form $\frac{a}{r}, a, ar$ is far cleaner.

   Tip

   Notice that the two sequences are geometric reversals: if $(1,2,4)$ has ratio $2$, then $(4,2,1)$ has ratio $\frac{1}{2}$. This symmetry always appears when the product condition gives $a^3$ and the sum of pairs is symmetric in $r$ and $1/r$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

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

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

   $b = \sqrt{2}\cdot(\sqrt{2})^2 = \sqrt{2}\cdot 2 = 2\sqrt{2}.$

   So the sequence $\sqrt{2},\ 2,\ 2\sqrt{2},\ 4$ is indeed a G.P. with common ratio $\sqrt{2}$, and the two inserted numbers in order are $2$ and $2\sqrt{2}$.

   ✓Final answer

   The correct option is (B) — $2,\ 2\sqrt{2}$.

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).
   • (D) $r=\tfrac{2\sqrt2}{3}$ — true (the other valid root).
   • (C) For $r=\tfrac13$: $S_\infty=\dfrac{4}{1-\tfrac13}=\dfrac{4}{2/3}=6$ — true.
   • (A) For $r=\tfrac{2\sqrt2}{3}$: $S_\infty=\dfrac{4}{1-\tfrac{2\sqrt2}{3}}=\dfrac{12}{3-2\sqrt2}=12(3+2\sqrt2)=36+24\sqrt2.$ Neither root yields $3+2\sqrt2$, so (A) is not true.

   ✓Final answer

   The statement that is not true is $S_\infty=3+2\sqrt2$ — option (A).

   ANSWER: A

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 Assuming $a \neq 0$ and $r^{2n} \neq 1$ (so the sums are defined), we cancel $a$ and the factor $r^{2n} - 1$:

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

   5. Solve for $r$ Note $r^2 - 1 = (r - 1)(r + 1)$. Multiply both sides by $(r - 1)(r + 1)$:

   $r + 1 = 5$

   Hence:

   $r = 4$

   Tip

   The number of terms $2n$ cancels out completely — the result does not depend on how many terms there are, as long as it’s even. That’s why the problem gives a unique answer.

   Watch out

   A common mistake is to forget that the odd-place terms form a GP with ratio $r^2$, not $r$. If you mistakenly use $r$ as the ratio for the odd terms, you’ll get a different (wrong) equation.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

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

      • If $r = 2$: from (1), $a(1 + 8) = 36 \Rightarrow 9a = 36 \Rightarrow a = 4$. The progression: $4, 8, 16, 32$.
      • If $r = \frac{1}{2}$: from (1), $a\left(1 + \frac{1}{8}\right) = 36 \Rightarrow \frac{9}{8}a = 36 \Rightarrow a = 32$. The progression: $32, 16, 8, 4$.

      Both are valid geometric progressions, just reversed. The options list only the increasing one: $4, 8, 16, 32$.

   Tip

   Notice that the two solutions are reverses of each other — the problem’s phrasing “first and last” picks the order, so only the increasing sequence matches the given options.

   Watch out

   A common mistake is to forget that $r$ could be fractional, giving a decreasing sequence. Always check both possibilities against the answer choices.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

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:

   $r = \sqrt[3]{-\frac{1}{8}} = -\frac{1}{2}.$

   (The cube root of a negative number is negative, and $\left(-\frac{1}{2}\right)^3 = -\frac{1}{8}$.)

   Watch out

   A common mistake is to forget the sign: since the 6th term is negative and the 9th term is positive, the ratio $r$ must be negative (an odd number of multiplications flips the sign). If you got $r = \frac{1}{2}$, check the sign of the ratio.

   Tip

   You don’t need to find the first term $a$ at all — dividing the two given terms directly gives $r^3$. This shortcut works whenever you know two terms and their positions.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

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:

   • (A) $a=1, r=-\frac{3}{4}$ — second term would be $-\frac{3}{4}$, not $\frac{3}{4}$.
   • (B) $a=3, r=\frac{1}{4}$ — matches our second solution.
   • (C) $a=-3, r=-\frac{1}{4}$ — second term would be $\frac{3}{4}$, but sum would be $\frac{-3}{1 - (-1/4)} = \frac{-3}{5/4} = -\frac{12}{5} \neq 4$.
   • (D) $a=-1, r=\frac{3}{4}$ — second term would be $-\frac{3}{4}$, not $\frac{3}{4}$.

   Only option (B) satisfies both conditions.

   Watch out

   A common mistake is to forget that the sum formula requires $|r| < 1$. Here both candidate $r$ values satisfy it, but always check.

   Tip

   You can also test each option directly: compute the sum and second term from the given $a$ and $r$ — only (B) gives both 4 and 3/4.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

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):

   $\left(\frac{3}{2}\right)^2 + 1 = 2\left(\frac{3}{2}\right) + \frac{1}{a}.$

   Compute:

   $\frac{9}{4} + 1 = 3 + \frac{1}{a} \quad \Rightarrow \quad \frac{13}{4} = 3 + \frac{1}{a}.$

   So:

   $\frac{13}{4} - 3 = \frac{1}{a} \quad \Rightarrow \quad \frac{1}{4} = \frac{1}{a}.$

   Hence $a = 4$.

   5. Determine all digits

      $a = 4$, $b = ar = 4 \times \frac{3}{2} = 6$, $c = ar^2 = 4 \times \frac{9}{4} = 9$.

      The number is 469.

   6. Sum of digits

      $4 + 6 + 9 = 19$.

   Watch out

   A common mistake is to forget that the hundreds digit cannot be zero, or to accept $r = -1$ which would give alternating signs and not a valid three-digit number with positive digits.

   Tip

   The second condition alone gave $r$ directly, independent of $a$. That made the problem simpler — always isolate the pure ratio condition first.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

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:

    $r+1 = 5 \quad \Rightarrow \quad r = 4$

    Watch out

    A common mistake is to forget that the odd‑placed terms themselves form a GP with ratio $r^2$, not $r$. Also, canceling $r^{2n}-1$ is valid only if it is nonzero — but if it were zero, the sums would be zero and the condition $0 = 5 \cdot 0$ would hold for any $r$, which is degenerate and not intended.

    Tip

    Notice that the number of terms $2n$ cancels out completely — the result is independent of how many terms there are, as long as it’s even.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

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

    $32^{6/5} = (2^5)^{6/5} = 2^{5 \cdot \frac{6}{5}} = 2^6 = 64$

    Watch out

    A common mistake is to think the product itself is infinite and therefore zero, but since each factor is greater than 1, the product grows — it does not shrink to zero. The exponent sum is finite, so the result is a finite number.

    Tip

    Whenever you see a repeated multiplication of the same base, immediately think: add the exponents. If the exponents form a geometric series, you can sum it in one step.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

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

    $P = 9^5$

    5. Express as a power of 3. Since $9 = 3^2$, we have $9^5 = (3^2)^5 = 3^{10}$.

    Watch out

    A common mistake is to think the product of five terms equals the cube of the third term, or to forget that the exponent on $r$ is the sum $0+1+2+3+4 = 10$, not $5$. Always check the exponent sum carefully.

    Tip

    For any G.P. with an odd number of terms $2n+1$, the product of all terms equals the middle term raised to the power $2n+1$. Here $n=2$, so product = (third term)$^5$.

    ✓Final answer

    The product is $3^{10}$, which corresponds to option (C).