Q.Find the sum to indicated number of terms in the geometric progression $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, \ldots$ $n$ terms.

Showing the 12 most recent of 43 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The sum of the first two terms of a geometric series is 12 and the third term is 16. Then the common ratio $r>0$ of the geometric progression is (A) $\dfrac{1}{2}$ (B) $\dfrac{2}{3}$ (C) 2 (D) 3 (E) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   From $a+ar=12$ and $ar^2=16$ derive $3r^2-4r-4=0$; the positive root is $r=2$.

   Let the GP have first term $a$ and ratio $r$. Then

   $a(1+r)=12,\qquad ar^2=16.$

   From the first, $a=\dfrac{12}{1+r}$. Substituting:

   $\frac{12r^2}{1+r}=16\Rightarrow 12r^2=16+16r\Rightarrow 3r^2-4r-4=0.$ …

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2. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The sum of four consecutive terms in a geometric progression is 960. If the fourth term is 8 times as large as the first term, then the smallest number in the geometric progression is (A) 40 (B) 64 (C) 128 (D) 160 (E) 240

   ›Reveal solutionSolution

   The ratio is $r=2$ (since term 4 $=8\times$ term 1), and $15a=960$ gives smallest term $a=64$.

   Let the four consecutive terms be $a,ar,ar^2,ar^3$. The condition $ar^3=8a$ gives $r^3=8$, so $r=2$.

   The sum is

   $a(1+r+r^2+r^3)=a(1+2+4+8)=15a=960\Rightarrow a=64.$ …

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3. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.A geometric series has common ratio $\frac{1}{3}$. If the sum of first four terms is 200, then the first term is (A) 90 (B) 100 (C) 115 (D) 145 (E) 135

   ›Reveal solutionSolution

   Solve $a\cdot\tfrac{40}{27}=200$ to get first term $135$.

   Geometric series with $r=\tfrac13$:

   $S_4=a\dfrac{1-r^4}{1-r}=a\dfrac{1-\tfrac1{81}}{1-\tfrac13}=a\cdot\dfrac{80/81}{2/3}=a\cdot\dfrac{40}{27}.$ …

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4. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The sum of the second and third terms of a G.P. is 8 and the fourth term is 4. The common ratio $r\neq 1$ is (A) $\dfrac{-1}{2}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{-1}{4}$ (D) $\dfrac{-1}{3}$ (E) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Ratio of $t_4$ to $(t_2+t_3)$ yields $2r^2-r-1=0$, so $r=-\tfrac12$.

   Let terms be $ar,ar^2,ar^3$. Given:

   $ar+ar^2=8,\qquad ar^3=4.$

   Divide the second by the first:

   $\dfrac{ar^3}{ar(1+r)}=\dfrac{r^2}{1+r}=\dfrac{4}{8}=\dfrac12.$ …

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5. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The first and the twentieth terms of a G.P. are 512 and $\dfrac{1}{1024}$ respectively. Then the common ratio is (A) $\dfrac{1}{2}$ (B) 2 (C) $\dfrac{1}{4}$ (D) 4 (E) $\dfrac{1}{8}$

   ›Reveal solutionSolution

   $r^{19}=2^{-19}$, so $r=\tfrac12$.

   With first term $a=512=2^9$ and $t_{20}=a r^{19}$:

   $2^9\,r^{19}=\dfrac{1}{1024}=2^{-10}\Rightarrow r^{19}=2^{-19}.$ …

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6. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.If $k$, $6$ and $k+5$ are the first three terms of a geometric series, then the possible values of the common ratio are (A) $\dfrac{1}{2},2$ (B) $2,-2$ (C) $\dfrac{1}{3},3$ (D) $\dfrac{3}{2},\dfrac{-2}{3}$ (E) $\dfrac{2}{3},\dfrac{-3}{2}$

   ›Reveal solutionSolution

   Solve $k^2+5k-36=0$ ($k=4,-9$), then $r=6/k=\tfrac32$ or $-\tfrac23$.

   For a GP, the middle term squared equals the product of neighbours:

   $6^2=k(k+5)\Rightarrow k^2+5k-36=0\Rightarrow(k-4)(k+9)=0.$ …

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7. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.The number of terms in the sequence $2, 6, 18, \ldots, 1458$ is (A) 14 (B) 12 (C) 10 (D) 8 (E) 7

   ›Reveal solutionSolution

   Geometric sequence with ratio 3; solve $2\cdot3^{k-1}=1458$.

   The sequence $2, 6, 18, \dots$ is a GP with first term $a=2$ and ratio $r=3$.

   The $k$-th term is $t_k = 2\cdot 3^{k-1}$. Set $2\cdot3^{k-1} = 1458$: …

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8. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.Let $t_1, t_2, t_3, \ldots, t_{2n - 2}, t_{2n - 1}, t_{2n}$ be in G.P. with common ratio $r$. Then (A) $t_1, t_3, t_5, \ldots, t_{2n - 5}, t_{2n - 3}, t_{2n - 1}$ are in G.P. with common ratio $r$ (B) $t_1, t_4, t_7, \ldots, t_{2n - 7}, t_{2n - 4}, t_{2n - 1}$ are in G.P. with common ratio $r^2$ (C) $t_1, t_3, t_5, \ldots, t_{2n - 5}, t_{2n - 3}, t_{2n - 1}$ are in G.P. with common ratio $r^2$ (D) $t_2, t_4, t_6, \ldots, t_{2n - 4}, t_{2n - 2}, t_{2n}$ are in G.P. with common ratio $r^3$ (E) $t_2, t_4, t_6, \ldots, t_{2n - 4}, t_{2n - 2}, t_{2n}$ are in G.P. with common ratio $r^5$

   ›Reveal solutionSolution

   Every-other term of a GP forms a GP with ratio $r^2$.

   In a GP $t_k=t_1 r^{k-1}$. The odd-indexed subsequence $t_1,t_3,t_5,\dots$ has

   $\dfrac{t_{k+2}}{t_k}=\dfrac{t_1 r^{k+1}}{t_1 r^{k-1}}=r^2.$ …

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9. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.If $\frac{4^{n + 1} + 16^{n + 1}}{4^n + 16^n}$ is the Geometric Mean between 4 and 16, then the value of $n$ is (A) $\frac{1}{2}$ (B) $\frac{3}{2}$ (C) 10 (D) $\frac{-1}{2}$ (E) 8

   ›Reveal solutionSolution

   The GM of 4 and 16 is 8; solving yields $n=-\tfrac12$.

   Geometric mean of 4 and 16 is $\sqrt{4\cdot16}=8$. So

   $\dfrac{4^{n+1}+16^{n+1}}{4^n+16^n}=\dfrac{4\cdot4^n+16\cdot16^n}{4^n+16^n}=8.$

   Let $a=4^n$, $b=16^n$: $4a+16b=8(a+b)\Rightarrow8b=4a\Rightarrow a=2b$, i.e. $4^n=2\cdot16^n$. …

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10. KEAM 2026Set eng-2026-04194 marksMCQ
    Q.The first and last term of a G.P. are 7 and 448 respectively. If the sum is 889, then the common ratio is (A) 4 (B) 2 (C) $\frac{1}{2}$ (D) $\frac{1}{4}$ (E) 3

    ›Reveal solutionSolution

    Solve $\dfrac{448r-7}{r-1}=889$ to get $r=2$.

    For a GP with first term $a=7$, last term $l=448$, sum $S=\dfrac{lr-a}{r-1}$:

    $\dfrac{448r-7}{r-1}=889.$ …

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11. KEAM 2026Set eng-2026-04204 marksMCQ
    Q.Let $t_1, t_2, t_3, \ldots, t_n$ be in G.P. Then $\left(\dfrac{t_4}{t_2}\right)^3$ is equal to (A) $\left(\dfrac{t_7}{t_3}\right)^2$ (B) $\left(\dfrac{t_7}{t_4}\right)^2$ (C) $\left(\dfrac{t_6}{t_4}\right)^2$ (D) $\left(\dfrac{t_7}{t_2}\right)^2$ (E) $\left(\dfrac{t_9}{t_8}\right)^2$

    ›Reveal solutionSolution

    For a geometric progression the ratio $t_m/t_n=r^{m-n}$. The given quantity is $(r^2)^3=r^6$; among the options only $(t_7/t_4)^2=(r^3)^2=r^6$ matches.

    Let the common ratio be $r$, so $t_k=t_1 r^{k-1}$ and $t_m/t_n=r^{m-n}$.

    Given quantity:

    $\left(\frac{t_4}{t_2}\right)^3=(r^{4-2})^3=(r^2)^3=r^6.$

    Evaluate each option:

    • (A) $(t_7/t_3)^2=(r^4)^2=r^8$
    • (B) $(t_7/t_4)^2=(r^3)^2=r^6$ ✓ …

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12. KEAM 2026Set eng-2026-04204 marksMCQ
    Q.The sum of the first $n$ terms in a G.P. is $s_n = 100 - 100^{1-n}$. The common ratio $r$ is (A) $\dfrac{3}{100}$ (B) $\dfrac{1}{500}$ (C) $\dfrac{1}{20}$ (D) $\dfrac{1}{50}$ (E) $\dfrac{1}{100}$

    ›Reveal solutionSolution

    A G.P. sum has the form $S_n=\frac{a}{1-r}(1-r^n)=\frac{a}{1-r}-\frac{a}{1-r}r^n$. Writing the given $S_n=100-100^{1-n}=100-100\cdot(1/100)^n$ shows the geometric factor is $r^n=(1/100)^n$, so $r=1/100$.

    For a G.P. with first term $a$ and ratio $r$,

    $S_n=\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}-\frac{a}{1-r}\,r^n.$

    The given sum is

    $S_n=100-100^{1-n}=100-100\cdot 100^{-n}=100-100\left(\tfrac{1}{100}\right)^n.$ …

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