ICSE Class 10 Geometric Progression — Mock Test (2027)
Free online mock test for Geometric Progression (ICSE Class 10 Mathematics) — 20 competency-based questions based on the latest CISCE 2027 syllabus, with instant marking. Try the samples below, then take the full test free.
What to expect: This mock test covers key concepts from the Geometric Progression chapter — including application-based and competency-focused questions aligned with how ICSE actually sets the paper.
Tip: Attempt without notes first to identify gaps, then review explanations for any wrong answers. Retake after a few days for best retention.
Sample questions
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1.The second term of a geometric progression (G.P.) is 9 and the sum of its infinite terms is 48. Which of the following represents the first three terms of the G.P.?
- A.16, 9, 81/16
- B.12, 9, 27/4
- C.6, 9, 27/2
- D.24, 9, 27/8
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2.If $a, b$ and $c$ are in G.P., which of the following proves that $\frac{1}{a + b}, \frac{1}{2b}$ and $\frac{1}{b + c}$ are in A.P.?
- A.$\frac{1}{a + b} + \frac{1}{b + c} = \frac{1}{b}$
- B.$\frac{1}{a + b} + \frac{1}{b + c} = \frac{2}{b}$
- C.$\frac{1}{a + b} \times \frac{1}{b + c} = \frac{1}{b^2}$
- D.$\frac{1}{a + b} - \frac{1}{b + c} = \frac{1}{b}$
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3.The $(p + q)^{\text{th}}$ term of a geometric progression (G.P.) is $m$ and its $(p - q)^{\text{th}}$ term is $n$. What is the $p^{\text{th}}$ term of the G.P.?
- A.$\sqrt{mn}$
- B.$\frac{m + n}{2}$
- C.$m + n$
- D.$\frac{m}{n}$
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4.Find the next three terms of the series: $\frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \dots$
- A.$2, 6, 18$
- B.$\frac{2}{1}, \frac{2}{0}, \text{undefined}$
- C.$6, 18, 54$
- D.$\frac{2}{81}, \frac{2}{243}, \frac{2}{729}$
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5.If for a geometric progression (G.P.), its $p^{\text{th}}$, $q^{\text{th}}$ and $r^{\text{th}}$ terms are $a$, $b$ and $c$ respectively, which of the following correctly proves the identity $a^{q - r} \cdot b^{r - p} \cdot c^{p - q} = 1$?
- A.$a^{q-r} \cdot b^{r-p} \cdot c^{p-q} = (AR^{p-1})^{q-r} \cdot (AR^{q-1})^{r-p} \cdot (AR^{r-1})^{p-q} = A^0 R^0 = 1$
- B.$a^{q-r} \cdot b^{r-p} \cdot c^{p-q} = A^{q-r + r-p + p-q} \cdot R^{(p-1)(q-r) + (q+1)(r-p) + (r-1)(p-q)} = A^0 R^{2(p+q+r)}$
- C.$a^{q-r} \cdot b^{r-p} \cdot c^{p-q} = (AR^{p+q+r})^0 = 1$ without further simplification
- D.$a^{q-r} \cdot b^{r-p} \cdot c^{p-q} = R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)} = R^{pqr}$
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