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Geometric Progression — ICSE Class 10 Maths Important Questions

14 ICSE Class 10 Maths practice questions on Geometric Progression, each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 30 September 2026

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Key concepts
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With answers

Geometric Progression — ICSE Class 10 Maths Important Questions

Common Ratio, Not Common Difference

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Quick answer

Paper-favourite ICSE Geometric Progression questions ask for the nnnth term Tn=ar n−1T_n=ar^{\,n-1}T_n=ar^\,n-1, the sum Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1}S_n=a(r^n-1)/r-1, which term equals a given value, and word problems on doubling, populations and salaries. Finding three numbers in GP from their sum and product appears very often.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Geometric Progression

In the ICSE Class 10 Maths chapter Geometric Progression a sequence has a constant ratio rrr between successive terms. You find any term with Tn=ar n−1T_n=ar^{\,n-1}T_n=ar^\,n-1, add terms using Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1}S_n=a(r^n-1)/r-1, and apply these to real situations such as populations, depreciation and compound growth. You'll frequently be asked to find a specific term of the sequence — the 7th term or the 10th term, say — or to find the nth term of a geometric progression given a couple of its terms.

General term $ar^{n-1}$Common ratioSum of $n$ terms of a GPNumbers in GP (sum and product)Word problems and growth

Key concepts & formulas

General term

For a GP with first term aaa and common ratio rrr, the nnnth term is Tn=ar n−1T_n=ar^{\,n-1}T_n=ar^\,n-1.

Sum of $n$ terms

Sn=a(rn−1)r−1S_n=\dfrac{a(r^n-1)}{r-1}S_n=a(r^n-1)/r-1 when r>1r>1r>1 and Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r}S_n=a(1-r^n)/1-r when r<1r<1r<1; for r=1r=1r=1, Sn=naS_n=naS_n=na.

Common ratio

r=Tn+1Tnr=\dfrac{T_{n+1}}{T_n}r=T_n+1T_n; a sequence is a GP only if this ratio is the same throughout.

Three numbers in GP

Take them as ar, a, ar\dfrac{a}{r},\,a,\,ara/r,\,a,\,ar so their product is a3a^3a^3, which simplifies sum-and-product problems.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The 555th term of the GP 2, 6, 18,…2,\,6,\,18,\dots2,\,6,\,18, is:

  1. (a)

    162162162

  2. (b)

    486486486

  3. (c)

    545454

  4. (d)

    969696

Show model answer

Answer: (a) 162162162.

Here a=2, r=3a=2,\ r=3a=2, r=3, so T5=ar4=2×34=2×81=162T_5=ar^{4}=2\times3^{4}=2\times81=162T_5=ar^4=2×3^4=2×81=162.

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Q2MCQModerate1 mark

In a GP the 333rd term is 121212 and the 666th term is 969696. The common ratio is:

  1. (a)

    222

  2. (b)

    333

  3. (c)

    444

  4. (d)

    12\tfrac1212

Show model answer

Answer: (a) 222.

T6T3=ar5ar2=r3=9612=8\dfrac{T_6}{T_3}=\dfrac{ar^{5}}{ar^{2}}=r^{3}=\dfrac{96}{12}=8T_6/T_3=ar^5ar^2=r^3=96/12=8, so r=2r=2r=2.

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Q3MCQEasy1 mark

Which term of the GP 3, 6, 12,…3,\,6,\,12,\dots3,\,6,\,12, is 384384384?

  1. (a)

    777th

  2. (b)

    888th

  3. (c)

    999th

  4. (d)

    101010th

Show model answer

Answer: (b) 888th.

arn−1=384⇒3×2 n−1=384⇒2 n−1=128=27ar^{n-1}=384\Rightarrow 3\times2^{\,n-1}=384\Rightarrow 2^{\,n-1}=128=2^{7}ar^n-1=384 3×2^\,n-1=384 2^\,n-1=128=2^7, so n−1=7, n=8n-1=7,\ n=8n-1=7, n=8.

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Q4MCQHOTS1 mark

The sum of the first 666 terms of the GP 1, −2, 4, −8,…1,\,-2,\,4,\,-8,\dots1,\,-2,\,4,\,-8, is:

  1. (a)

    −21-21-21

  2. (b)

    212121

  3. (c)

    −63-63-63

  4. (d)

    636363

Show model answer

Answer: (a) −21-21-21.

a=1, r=−2a=1,\ r=-2a=1, r=-2. S6=a(r6−1)r−1=1(64−1)−2−1=63−3=−21S_6=\dfrac{a(r^{6}-1)}{r-1}=\dfrac{1(64-1)}{-2-1}=\dfrac{63}{-3}=-21S_6=a(r^6-1)r-1=1(64-1)/-2-1=63/-3=-21.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The sequence 5, 10, 20, 40,…5,\,10,\,20,\,40,\dots5,\,10,\,20,\,40, is a GP with common ratio 222.

Reason (R): A sequence is a GP if the ratio of every term to its preceding term is constant.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (a) Each ratio 105=2010=4020=2\dfrac{10}{5}=\dfrac{20}{10}=\dfrac{40}{20}=210/5=20/10=40/20=2 is constant, so it is a GP with r=2r=2r=2, and R correctly explains A.

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Very short answer questions (2 marks)

Q6Very ShortModerate2 marks

Find the value(s) of xxx for which x+9x+9x+9, x−6x-6x-6 and 444 are three consecutive terms of a GP.

Show model answer

For three consecutive terms of a GP, (middle term)2=(\text{middle term})^2=(middle term)^2= product of the other two:

(x−6)2=4(x+9)⇒x2−12x+36=4x+36⇒x2−16x=0(x-6)^2=4(x+9)\Rightarrow x^2-12x+36=4x+36\Rightarrow x^2-16x=0(x-6)^2=4(x+9) x^2-12x+36=4x+36 x^2-16x=0, so x=0x=0x=0 or x=16x=16x=16.

(Check: x=0x=0x=0 gives 9, −6, 49,\,-6,\,49,\,-6,\,4 with r=−23r=-\tfrac{2}{3}r=-23; x=16x=16x=16 gives 25, 10, 425,\,10,\,425,\,10,\,4 with r=25r=\tfrac{2}{5}r=25. Both are GPs.)

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Q7Very ShortModerate2 marks

Find the sum: 1+3+9+⋯+21871+3+9+\dots+21871+3+9++2187.

Show model answer

a=1, r=3a=1,\ r=3a=1, r=3. The last term 2187=37=arn−12187=3^{7}=ar^{n-1}2187=3^7=ar^n-1 gives n=8n=8n=8.

S8=a(rn−1)r−1=1(38−1)3−1=6561−12=65602=3280.S_8=\dfrac{a(r^{n}-1)}{r-1}=\dfrac{1(3^{8}-1)}{3-1}=\dfrac{6561-1}{2}=\dfrac{6560}{2}=3280.S_8=a(r^n-1)r-1=1(3^8-1)3-1=6561-1/2=6560/2=3280.

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Q8Very ShortEasy2 marks

Find the sum of the first 666 terms of the GP 64, 32, 16,…64,\,32,\,16,\dots64,\,32,\,16,

Show model answer

a=64, r=12a=64,\ r=\tfrac12a=64, r=12.

S6=a(1−r6)1−r=64(1−164)12=128×6364=126.S_6=\dfrac{a(1-r^{6})}{1-r}=\dfrac{64\left(1-\tfrac{1}{64}\right)}{\tfrac12}=128\times\dfrac{63}{64}=126.S_6=a(1-r^6)1-r=64(1-164)12=128×63/64=126.

(Check: 64+32+16+8+4+2=12664+32+16+8+4+2=12664+32+16+8+4+2=126.)

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Short answer questions (3 marks)

Q9Short AnswerModerate3 marks

The 444th term of a GP is 242424 and its 777th term is 192192192. Find the GP.

Show model answer

T4=ar3=24T_4=ar^{3}=24T_4=ar^3=24 and T7=ar6=192T_7=ar^{6}=192T_7=ar^6=192.

Dividing, r3=19224=8⇒r=2r^{3}=\dfrac{192}{24}=8\Rightarrow r=2r^3=192/24=8 r=2. Then a×8=24⇒a=3a\times8=24\Rightarrow a=3a×8=24 a=3.

The GP is 3, 6, 12, 24,…3,\,6,\,12,\,24,\dots3,\,6,\,12,\,24,

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Q10Short AnswerModerate3 marks

How many terms of the GP 2, 4, 8,…2,\,4,\,8,\dots2,\,4,\,8, add up to 510510510?

Show model answer

a=2, r=2a=2,\ r=2a=2, r=2.

Sn=a(rn−1)r−1=2(2n−1)2−1=2(2n−1)=510.S_n=\dfrac{a(r^{n}-1)}{r-1}=\dfrac{2(2^{n}-1)}{2-1}=2(2^{n}-1)=510.S_n=a(r^n-1)r-1=2(2^n-1)2-1=2(2^n-1)=510.

2n−1=255⇒2n=256=28⇒n=8.2^{n}-1=255\Rightarrow 2^{n}=256=2^{8}\Rightarrow n=8.2^n-1=255 2^n=256=2^8 n=8.

So 888 terms are needed.

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Q11Short AnswerHOTS3 marks

The sum of three numbers in GP is 212121 and their product is 216216216. Find the numbers.

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Let the numbers be ar, a, ar\dfrac{a}{r},\,a,\,ara/r,\,a,\,ar.

Product: ar×a×ar=a3=216⇒a=6.\dfrac{a}{r}\times a\times ar=a^{3}=216\Rightarrow a=6.a/r× a× ar=a^3=216 a=6.

Sum: 6(1r+1+r)=21⇒1r+r=216−1=52.6\left(\dfrac1r+1+r\right)=21\Rightarrow \dfrac1r+r=\dfrac{21}{6}-1=\dfrac{5}{2}.6(1r+1+r)=21 1r+r=21/6-1=5/2.

2r2−5r+2=0⇒(2r−1)(r−2)=0⇒r=2 or 12.2r^{2}-5r+2=0\Rightarrow (2r-1)(r-2)=0\Rightarrow r=2\ \text{or}\ \tfrac12.2r^2-5r+2=0 (2r-1)(r-2)=0 r=2 or 12.

The numbers are 3, 6, 12.3,\,6,\,12.3,\,6,\,12.

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Long answer questions (5 marks)

Q12Long AnswerModerate5 marks

In a GP the 222nd term is 999 and the 444th term is 818181 (all terms positive). Find (i) the common ratio, (ii) the first term, and (iii) the sum of the first 666 terms.

Show model answer

(i) T2=ar=9T_2=ar=9T_2=ar=9 and T4=ar3=81T_4=ar^{3}=81T_4=ar^3=81.

ar3ar=r2=819=9⇒r=3\dfrac{ar^{3}}{ar}=r^{2}=\dfrac{81}{9}=9\Rightarrow r=3ar^3ar=r^2=81/9=9 r=3 (positive).

(ii) ar=9⇒a×3=9⇒a=3.ar=9\Rightarrow a\times3=9\Rightarrow a=3.ar=9 a×3=9 a=3.

(iii) S6=a(r6−1)r−1=3(36−1)3−1=3(729−1)2=3×7282=1092.S_6=\dfrac{a(r^{6}-1)}{r-1}=\dfrac{3(3^{6}-1)}{3-1}=\dfrac{3(729-1)}{2}=\dfrac{3\times728}{2}=1092.S_6=a(r^6-1)r-1=3(3^6-1)3-1=3(729-1)/2=3×728/2=1092.

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Q13Long AnswerModerate5 marks

A car costing Rs 5,00,0005{,}00{,}0005,00,000 loses 20%20\%20\% of its value every year.

(i) Show that its values at the end of successive years form a GP, and write the common ratio.

(ii) Find its value at the end of 444 years.

(iii) After how many complete years does its value first fall below Rs 2,00,0002{,}00{,}0002,00,000?

Show model answer

(i) Each year the value becomes 80%80\%80\% of the previous year's value, so each value is 0.80.80.8 times the one before. The values Rs 4,00,0004{,}00{,}0004,00,000, Rs 3,20,0003{,}20{,}0003,20,000, Rs 2,56,000,…2{,}56{,}000,\dots2,56,000, form a GP with r=0.8=45r=0.8=\tfrac45r=0.8=45.

(ii) Value after nnn years =5,00,000×(0.8)n=5{,}00{,}000\times(0.8)^{n}=5,00,000×(0.8)^n. After 444 years: 5,00,000×0.4096=5{,}00{,}000\times0.4096=5,00,000×0.4096= Rs 2,04,8002{,}04{,}8002,04,800.

(iii) After 444 years the value is Rs 2,04,8002{,}04{,}8002,04,800, still above Rs 2,00,0002{,}00{,}0002,00,000. After 555 years it is 2,04,800×0.8=2{,}04{,}800\times0.8=2,04,800×0.8= Rs 1,63,8401{,}63{,}8401,63,840. So the value first falls below Rs 2,00,0002{,}00{,}0002,00,000 after 555 years.

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Case-based questions (4 marks)

Q14Case-basedEasy4 marks

A student starts a 101010-day saving challenge. On day 111 she saves Rs 111, on day 222 Rs 222, on day 333 Rs 444, and the amount doubles each day.

(i) How much does she save on day 555?

(ii) How much does she save on day 101010?

(iii) What is her total saving over the 101010 days?

(iv) On which day does she first save more than Rs 200200200?

Show model answer

The daily amounts form a GP with a=1, r=2a=1,\ r=2a=1, r=2.

(i) Day 555: T5=ar4=1×24=T_5=ar^{4}=1\times2^{4}=T_5=ar^4=1×2^4= Rs 16.16.16.

(ii) Day 101010: T10=ar9=29=T_{10}=ar^{9}=2^{9}=T_10=ar^9=2^9= Rs 512.512.512.

(iii) Total: S10=a(r10−1)r−1=1(210−1)2−1=1023S_{10}=\dfrac{a(r^{10}-1)}{r-1}=\dfrac{1(2^{10}-1)}{2-1}=1023S_10=a(r^10-1)r-1=1(2^10-1)2-1=1023, i.e. Rs 1023.1023.1023.

(iv) Tn=2 n−1T_n=2^{\,n-1}T_n=2^\,n-1. T8=27=T_8=2^{7}=T_8=2^7= Rs 128128128 and T9=28=T_9=2^{8}=T_9=2^8= Rs 256256256, so she first saves more than Rs 200200200 on day 999.

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All ICSE Class 10 Maths Chapters

GST (Goods and Services Tax)Banking (Recurring Deposit Accounts)Shares and DividendsLinear Inequations (In One Variable)Quadratic EquationsSolving (Simple) Problems Based on Quadratic EquationsRatio and ProportionRemainder and Factor TheoremsMatricesArithmetic ProgressionGeometric ProgressionReflectionSection and Mid-Point FormulaEquation of a LineSimilarity (With Applications to Maps and Models)Loci (Locus and Its Constructions)CirclesTangents and Intersecting ChordsConstructions (Circles)Cylinder, Cone and Sphere (Surface Area and Volume)Trigonometrical IdentitiesHeights and DistancesGraphical Representation (Histograms and Ogives)Measures of Central Tendency (Mean, Median, Quartiles and Mode)Probability

Frequently asked questions

  • Do these Geometric Progression questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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