MathsICSE Class 10 · Chapter 9

Arithmetic and Geometric Progression ICSE Class 10: General Term, Sum and Applications

Q: Is the sum of an infinite G.P. in the ICSE Class 10 syllabus?

The 2027 syllabus says 'finding sum of their first n terms', so practise sums of a finite number of terms.

Q: Which sum formula should I use for an A.P.?

If you know the last term, use n/2(a + l). Otherwise use n/2[2a + (n − 1)d]. Both give the same answer.

Q: Why can two values of n give the same A.P. sum?

If the A.P. goes from positive to negative terms, the extra negative terms can cancel some positive ones, so the running total returns to the same value. Both values are valid unless the question rules one out.

Q: How do I find n when the G.P. term is a big number?

Divide by a, then write the result as a power of r (for example 729 = 3⁶). Knowing powers of 2, 3, 4 and 5 makes this quick.

Updated 14 October 2026 · Fact-checked

An arithmetic progression (A.P.) adds the same number d each time; a geometric progression (G.P.) multiplies by the same number r. ICSE Class 10 asks for the general term, a + (n − 1)d or arⁿ⁻¹, the sum of the first n terms, and simple applications such as savings, seating and growth.

What Arithmetic and Geometric Progression is about

A sequence is a list of numbers in order. Two kinds matter here.

In an arithmetic progression (A.P.), each term is the previous one plus a fixed number, the common difference $d$ : for example 7, 11, 15, 19, … ( $d = 4$ ).

In a geometric progression (G.P.), each term is the previous one times a fixed number, the common ratio $r$ : for example 3, 6, 12, 24, … ( $r = 2$ ).

The CISCE 2027 syllabus asks you to find the general ( $n$ th) term and the sum of the first $n$ terms of each, with simple applications. The examiners' reports for 2024 and 2025 list 'Tn and Sn of A.P. & G.P.' among the topics candidates found difficult. The errors they describe include mixing up the A.P. and G.P. formulas, writing $a \times r^{n - 1}$ as $(a r)^{n - 1}$ (for example $7 \times 3^{n - 1}$ as $2 1^{n - 1}$ ), using $n$ instead of $n - 1$ in the general term, and confusing the last term with the sum.

Finding $n$ from a sum often gives a quadratic equation, and finding $n$ from a G.P. term needs indices, so revise both.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Arithmetic and Geometric Progression is in the 2. Algebra (vi) Arithmetic and Geometric Progression part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

A.P.: common difference and general term

In an A.P. with first term $a$ and common difference $d$ , the terms are $a, a + d, a + 2 d, \dots$ . The $n$ th term (general term) is
$T_{n} = a + (n - 1) d .$
The multiplier is $(n - 1)$ , not $n$ : to reach the 7th term you add $d$ six times. The common difference can be negative or a fraction: $d = T_{2} - T_{1}$ (any term minus the one before it).

If the last term is $l$ and you want the $n$ th term from the end, count backwards: $l - (n - 1) d$ .

Worked example

Find the 20th term of the A.P. 7, 11, 15, …

Try it first, then show the solutionSolution

$a = 7$ , $d = 4$ .
$T_{20} = 7 + 19 \times 4 = 83$ .

Answer: 83

A.P. from two given terms

If two terms are given, write each with the general-term formula. Subtracting the two equations removes $a$ and gives $d$ . Then find $a$ , and any term you need.

Worked example

The 5th term of an A.P. is 19 and the 12th term is 47. Find the 30th term.

Try it first, then show the solutionSolution

$a + 4 d = 19$ and $a + 11 d = 47$ .
Subtract: $7 d = 28$ , so $d = 4$ and $a = 3$ .
$T_{30} = 3 + 29 \times 4 = 119$ .

Answer: 119

Sum of the first n terms of an A.P.

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l),$
where $l = T_{n}$ is the last term. Use the second form whenever the last term is known; it is shorter and safer.

Useful link: $T_{n} = S_{n} - S_{n - 1}$ . So if a question gives a formula for $S_{n}$ , you can find any term from it.

Worked example

Find the sum of the first 20 terms of 3, 8, 13, …

Try it first, then show the solutionSolution

$a = 3$ , $d = 5$ , $n = 20$ .
$S_{20} = 10 [6 + 19 \times 5] = 10 \times 101 = 1010$ .

Answer: 1010

Finding n from a sum

If the sum is given and $n$ is unknown, substitute into $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . This gives a quadratic in $n$ . Solve it and keep only positive whole-number values. Sometimes both roots are valid: in an A.P. that becomes negative, the extra negative terms can cancel some positive ones and give the same sum again.

Worked example

How many terms of 2, 6, 10, … must be taken to give a sum of 450?

Try it first, then show the solutionSolution

$S_{n} = \frac{n}{2} [4 + 4 (n - 1)] = 2 n^{2}$ .
$2 n^{2} = 450 \Rightarrow n^{2} = 225 \Rightarrow n = 15$ .

Answer: 15 terms

G.P.: common ratio and general term

In a G.P. with first term $a$ and common ratio $r$ , the terms are $a, a r, a r^{2}, \dots$ , and
$T_{n} = a r^{n - 1} .$
Find $r$ by dividing any term by the one before it: $r = \frac{T _{2}}{T _{1}}$ . The ratio can be negative (signs alternate) or a fraction (terms shrink).

Only $r$ is raised to the power: $5 \times 3^{n - 1}$ is not $1 5^{n - 1}$ .

Worked example

Find the 8th term of the G.P. 3, 6, 12, …

Try it first, then show the solutionSolution

$a = 3$ , $r = 2$ .
$T_{8} = 3 \times 2^{7} = 3 \times 128 = 384$ .

Answer: 384

Sum of the first n terms of a G.P.

$S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} (r > 1), S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} (r < 1) .$
The two forms are the same formula; choose the one that keeps the numbers positive. (If $r = 1$ , every term is $a$ and $S_{n} = na$ .)

The power in the sum is $r^{n}$ , while the power in the $n$ th term is $r^{n - 1}$ . Examiners noted candidates writing $2^{n - 1}$ where $2^{n} - 1$ was needed.

Worked example

Find the sum of the first 6 terms of 2, 6, 18, … and of the first 7 terms of 64, 32, 16, …

Try it first, then show the solutionSolution

First: $a = 2$ , $r = 3$ : $S_{6} = \frac{2 ( 3 ^{6} - 1 )}{2} = 729 - 1 = 728$ .
Second: $a = 64$ , $r = \frac{1}{2}$ : $S_{7} = \frac{64 ( 1 - \frac{1}{128} )}{\frac{1}{2}} = 128 \times \frac{127}{128} = 127$ .

Answer: 728 and 127

Choosing terms cleverly, and simple applications

When three numbers in A.P. have a known sum, take them as $a - d$ , $a$ , $a + d$ : the sum is $3 a$ , so $a$ comes out at once. For three numbers in G.P. with a known product, take $\frac{a}{r}$ , $a$ , $a r$ : the product is $a^{3}$ .

In applications, decide first whether the quantity goes up by a fixed amount (A.P.: savings with a fixed increase, seats added per row) or by a fixed factor (G.P.: doubling, tripling, a fixed fraction each time). Then decide whether the question wants one term (the amount in a particular month or row) or a sum (the total so far).

Worked example

Three numbers in A.P. have sum 21 and product 231. Find them.

Try it first, then show the solutionSolution

Let them be $a - d$ , $a$ , $a + d$ : $3 a = 21$ , $a = 7$ .
$7 (49 - d^{2}) = 231 \Rightarrow 49 - d^{2} = 33 \Rightarrow d = \pm 4$ .
The numbers are 3, 7, 11 (or 11, 7, 3).

Answer: 3, 7, 11

Maths in real life: Arithmetic and Geometric Progression

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Sum of an A.P.

Suppose you decide to save ₹50 in the first week and ₹10 more than the previous week every week after that.

How much will you have saved after 20 weeks, and how much do you save in the 20th week?

Try it first, then show the solutionSolution

Weekly savings form an A.P. with $a = 50$ , $d = 10$ .
20th week: $T_{20} = 50 + 19 \times 10 = ₹240$ .
Total: $S_{20} = \frac{20}{2} (50 + 240) = 10 \times 290 = ₹2, 900$ .

Answer: ₹2,900 in all; ₹240 in week 20

Memory hook: Fixed extra each week: an A.P.

Sum of a G.P.

Imagine a school video gets 200 views on day 1, and each day it gets three times as many new views as the day before.

How many views does it have in total after 5 days?

Try it first, then show the solutionSolution

New views form a G.P. with $a = 200$ , $r = 3$ .
$S_{5} = \frac{200 ( 3 ^{5} - 1 )}{3 - 1} = 100 \times 242 = 24, 200$ .

Answer: 24,200 views

Memory hook: Fixed factor each day: a G.P.

A.P. with a negative common difference

Suppose logs are stacked in rows, with 20 logs in the bottom row, 19 in the next, 18 in the next, and so on, up to 5 logs in the top row.

How many rows are there and how many logs in all?

Try it first, then show the solutionSolution

$a = 20$ , $d = - 1$ , last term 5: $5 = 20 + (n - 1) (- 1) \Rightarrow n = 16$ .
Total $= \frac{16}{2} (20 + 5) = 8 \times 25 = 200$ .

Answer: 16 rows; 200 logs

Memory hook: Rows shrink by one: an A.P. going down.

Topics in this chapter

Arithmetic and Geometric Progression formula sheet

Common difference (A.P.): $d = T_{n} - T_{n - 1}$; Can be negative or a fraction.
General term (A.P.): $T_{n} = a + (n - 1) d$; (n − 1), not n.
Sum of n terms (A.P.): $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l)$; Use (a + l) when the last term is known.
Term from the sum: $T_{n} = S_{n} - S_{n - 1}$; For any sequence.
Common ratio (G.P.): $r = \frac{T _{n}}{T _{n - 1}}$; Can be negative or a fraction.
General term (G.P.): $T_{n} = a r^{n - 1}$; Only r is raised to the power.
Sum of n terms (G.P.): $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} (r > 1); S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} (r < 1)$; Power n in the sum, n − 1 in the term.
Three terms: $A.P.: a - d, a, a + d; G.P.: \frac{a}{r}, a, a r$; Makes the sum (A.P.) or product (G.P.) simple.

How the board asks Arithmetic and Geometric Progression

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
MCQ: a term of an A.P. or G.P.	Section A, Question 1: the 7th or 10th term, sometimes with algebraic terms like 1/a, 1/a + 1, …	Find d or r first; use (n − 1) as the multiplier or power.
General term given, find a, d and a sum	For example: 'The nth term of an A.P. is Tn = 4(9 − n). Find its first term, common difference and the sum of its first 20 terms.'	Put n = 1 and n = 2 to get a and T₂; d = T₂ − T₁; then the sum formula, watching signs.
First term, last term and sum given	Find the number of terms and the common difference.	Use $S_{n} = \frac{n}{2} (a + l)$ for n first, then $l = a + (n - 1) d$ for d.
Which term of a G.P. is a given number?	For example: 'If 2430 is the nth term of the G.P. 10, 30, 90, …, find n and the sum of the first n terms.'	$a r^{n - 1} =$ given number; divide by a; write the result as a power of r; then $S_{n}$ .
How many terms give a sum?	A sum of an A.P. or G.P. is given; find n.	A.P.: quadratic in n; G.P.: solve $r^{n} = \dots$ using powers.
Simple applications	Savings, seats in rows, logs in a pile, a ball bouncing, messages spreading.	Decide A.P. or G.P.; decide term or sum; answer in context.

Tricks to remember Arithmetic and Geometric Progression

Short memory aids. Say each one aloud once, then use it on a question the same day.

A.P. Adds, G.P. Grows by multiplying
Arithmetic: Add the same number. Geometric: multiply by the same number.
Use it when: Spotting the type before choosing a formula.
One less step than the term number
To reach the nth term you take (n − 1) steps from the first: (n − 1)d or r to the power (n − 1).
Use it when: General-term formulas.
Only r wears the power
In arⁿ⁻¹ the power sits on r alone; a stays outside.
Use it when: Avoiding 21ⁿ⁻¹ for 7 × 3ⁿ⁻¹.
First plus last, times half the count
Sₙ = n/2 × (first + last): pair the first and last terms, the second and second-last, and so on.
Use it when: A.P. sums when l is known.

Arithmetic and Geometric Progression: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: A term with negative d · 1 mark · MCQBoard level

The 10th term of the A.P. 5, 2, −1, … is:
(a) −22
(b) −25
(c) 32
(d) −19

Try it first, then show the solutionSolution

$d = - 3$ ; $T_{10} = 5 + 9 (- 3) = - 22$ .

Answer: (a)

Example 2: Recognising a G.P. · 1 mark · Assertion-reasonBoard level

Assertion (A): 3, 6, 12, 24 is a G.P. with common ratio 2.
Reason (R): In a G.P., the ratio of every term to the term before it is the same.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Try it first, then show the solutionSolution

$\frac{6}{3} = \frac{12}{6} = \frac{24}{12} = 2$ : constant ratio, so (A) is true by (R).

Answer: (c)

Example 3: From the general term to a sum · 3 marks · Short answerHarder board level

The $n$ th term of an A.P. is $T_{n} = 5 - 3 n$ . Find (a) its first term and common difference, (b) the sum of its first 20 terms.

Try it first, then show the solutionSolution

$T_{1} = 2$ , $T_{2} = - 1$ , so $a = 2$ , $d = - 3$ .
$S_{20} = 10 [4 + 19 (- 3)] = 10 (4 - 57) = - 530$ .

Answer: (a) a = 2, d = −3 (b) −530

Example 4: First term, last term and sum · 4 marks · Short answerHarder board level

The first term of an A.P. is 7, the last term is 97 and the sum of all its terms is 832. Find the number of terms and the common difference.

Try it first, then show the solutionSolution

$S_{n} = \frac{n}{2} (a + l)$ : $832 = \frac{n}{2} (104) = 52 n$ , so $n = 16$ .
$l = a + (n - 1) d$ : $97 = 7 + 15 d$ , so $d = 6$ .

Answer: 16 terms; d = 6

Example 5: Which term of a G.P.? · 4 marks · Short answerHarder board level

Which term of the G.P. 5, 15, 45, … is 3645? Find the sum of the G.P. up to that term.

Try it first, then show the solutionSolution

$5 \times 3^{n - 1} = 3645 \Rightarrow 3^{n - 1} = 729 = 3^{6}$ .
$n - 1 = 6$ , so $n = 7$ .
$S_{7} = \frac{5 ( 3 ^{7} - 1 )}{3 - 1} = \frac{5 \times 2186}{2} = 5465$ .

Answer: 7th term; sum 5465

Example 6: Two-digit multiples of 7 · 3 marks · Long answerBoard level

Find the sum of all two-digit natural numbers that are multiples of 7.

Try it first, then show the solutionSolution

They are 14, 21, …, 98: $a = 14$ , $d = 7$ , $l = 98$ .
$98 = 14 + (n - 1) 7 \Rightarrow n = 13$ .
$S = \frac{13}{2} (14 + 98) = 13 \times 56 = 728$ .

Answer: 728

Example 7: Steady growth or doubling? · 4 marks · Long answerHOTS / competency

Riya saves ₹500 in month 1 and ₹100 more each month after that. Her brother saves ₹100 in month 1 and doubles his saving each month. (a) Write each person's total after $n$ months. (b) Find the first month at the end of which the brother's total is more than Riya's. (c) Explain why the doubling plan must overtake any plan that adds a fixed amount.

Try it first, then show the solutionSolution

(a) Riya: A.P., $S_{n} = \frac{n}{2} [1000 + 100 (n - 1)] = 50 n (n + 9)$ . Brother: G.P., $S_{n} = 100 (2^{n} - 1)$ .
(b) $n = 4$ : 2600 vs 1500; $n = 5$ : 3500 vs 3100; $n = 6$ : 4500 vs 6300. First at the end of month 6.
(c) Riya's monthly saving grows by ₹100 each time, but the brother's doubles, so his monthly saving eventually exceeds hers by more and more; once that happens his total catches up and passes hers.

Answer: (a) 50n(n + 9) and 100(2ⁿ − 1) (b) month 6 (c) multiplying grows faster than adding.

Example 8: Terms from a sum formula · 4 marks · Long answerHarder board level

The sum of the first $n$ terms of an A.P. is $S_{n} = 3 n^{2} + 5 n$ . Find the first term, the common difference and the 10th term.

Try it first, then show the solutionSolution

$T_{1} = S_{1} = 8$ .
$T_{2} = S_{2} - S_{1} = 22 - 8 = 14$ , so $d = 6$ .
$T_{10} = 8 + 9 \times 6 = 62$ (check: $S_{10} - S_{9} = 350 - 288 = 62$ ).

Answer: a = 8, d = 6, T₁₀ = 62

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Seating in a school auditorium

A school in Bhubaneswar is building an auditorium. The first row will have 20 seats and each row after it will have 2 more seats than the row in front.

(i) · 3 marks
How many seats will the 25th row have?
Answer
Answer: 68
$T_{25} = 20 + 24 \times 2 = 68$ .
(ii) · 3 marks
Find the total number of seats in 25 rows.
Answer
Answer: 1100
$S_{25} = \frac{25}{2} (20 + 68) = 25 \times 44 = 1100$ .
(iii) · 4 marks
The school needs at least 1,500 seats. What is the least number of rows needed, and how many seats will there then be?
Answer
Answer: 31 rows; 1550 seats
$S_{n} = \frac{n}{2} [40 + 2 (n - 1)] = n (n + 19)$ . $S_{30} = 30 \times 49 = 1470 < 1500$ and $S_{31} = 31 \times 50 = 1550 \geq 1500$ . (Solving $n^{2} + 19 n - 1500 = 0$ gives $n \approx 30.4$ , so round up.) 31 rows, 1550 seats.

News spreading through a town

News of a school event spreads through a town. On day 1, 4 people hear it. On each later day, the number of new people who hear it is three times the number of new people on the day before.

(i) · 3 marks
How many new people hear the news on day 6?
Answer
Answer: 972
G.P. with $a = 4$ , $r = 3$ : $T_{6} = 4 \times 3^{5} = 972$ .
(ii) · 3 marks
How many people in total have heard it by the end of day 6?
Answer
Answer: 1456
$S_{6} = \frac{4 ( 3 ^{6} - 1 )}{2} = 2 \times 728 = 1456$ .
(iii) · 4 marks
On which day does the total first exceed 10,000? Give the total at the end of that day.
Answer
Answer: Day 8; 13,120
$S_{n} = 2 (3^{n} - 1) > 10000 \Rightarrow 3^{n} > 5001$ . $3^{7} = 2187$ , $3^{8} = 6561$ , so $n = 8$ . $S_{8} = 2 \times 6560 = 13, 120$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the 12th term and the sum of the first 12 terms of the A.P. whose first term is 8 and common difference is 3. (3 marks)

Step	Marks
$T_{12} = 8 + 11 \times 3 = 41$	1
$S_{12} = \frac{12}{2} (8 + 41)$	1
$= 6 \times 49 = 294$	1

The first term of an A.P. is 4, the last term is 64 and the sum is 544. Find the number of terms and the common difference. (4 marks)

Step	Marks
$544 = \frac{n}{2} (4 + 64)$	1
$n = 16$	1
$64 = 4 + 15 d$	1
$d = 4$	1

Which term of the G.P. 6, 18, 54, … is 4374? Hence find the sum of the G.P. up to that term. (5 marks)

Step	Marks
$a = 6$ , $r = 3$ ; $6 \times 3^{n - 1} = 4374$	1
$3^{n - 1} = 729 = 3^{6}$	1
$n = 7$	1
$S_{7} = \frac{6 ( 3 ^{7} - 1 )}{3 - 1}$	1
$= 3 \times 2186 = 6558$	1

Where students lose marks in Arithmetic and Geometric Progression

The slips examiners see most often, and what to write instead.

Mistake: Using n instead of n − 1 in the general term
CISCE examiners' comments, 2024: for the 7th term of an A.P. with d = 1, candidates added 7 instead of 6, or used a wrong formula.
Wrong working
$T_{7} = \frac{1}{a} + 7$
Right working
$T_{7} = \frac{1}{a} + 6 \times 1 = \frac{1}{a} + 6$
Do this: Write $T_{n} = a + (n - 1) d$ and substitute $n - 1 = 6$ on its own line.
Usually costs: 1 mark.
Mistake: Multiplying a into the power in a G.P.
CISCE examiners' comments, 2024 and 2025: candidates wrote $15 \times 2^{n - 1}$ as $3 0^{n - 1}$ , and $7 \times 3^{n - 1}$ as $2 1^{n - 1}$ .
Do this: Only r is raised to the power. Divide by a first: $7 \times 3^{n - 1} = 1701 \Rightarrow 3^{n - 1} = 243$ .
Usually costs: Most of the question.
Mistake: Mixing up A.P. and G.P. formulas, or term and sum
CISCE examiners' comments, 2024 and 2025: candidates used A.P. formulas for a G.P., equated the nth term to a given sum, or wrote $2^{n - 1}$ instead of $2^{n} - 1$ .
Do this: First write 'This is an A.P./G.P. with a = …, d/r = …'. Then ask: does the question give a term or a sum?
Usually costs: Usually all the marks for that part.
Mistake: Confusing the last term with the sum
CISCE examiners' comments, 2025: given first term 5, last term 50 and sum 440, candidates took n as 50 and the nth term as 440.
Do this: Label each given number: a = …, l = …, Sₙ = … before using any formula.
Usually costs: Both parts of a 3-mark question.
Mistake: Not finding a and d from a general term
CISCE examiners' comments, 2024: given $T_{n} = 6 (7 - n)$ , candidates did not substitute n = 1 and n = 2, and got the sum's sign wrong (900 instead of −900).
Do this: Write out $T_{1}$ , $T_{2}$ and $T_{3}$ explicitly, then $d = T_{2} - T_{1}$ .
Usually costs: 1 to 2 marks.
Mistake: Wrong powers of small numbers
CISCE examiners' comments, 2025: candidates wrote $3^{6} = 243$ instead of 729, and took r as 1/3 instead of 3.
Do this: Keep a list: powers of 2 to 2¹⁰, of 3 to 3⁸, of 4 and 5 to the 5th power. r = later term ÷ earlier term.
Usually costs: 1 to 2 marks.

Practice questions on Arithmetic and Geometric Progression

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The common difference of the A.P. $\frac{1}{3}, \frac{5}{6}, \frac{4}{3}, \dots$ is:
(a) $\frac{1}{3}$
(b) $\frac{1}{2}$
(c) $\frac{2}{3}$
(d) $\frac{1}{6}$
Hint
Subtract the first term from the second.
Answer
(b) $\frac{5}{6} - \frac{2}{6} = \frac{1}{2}$
Q2 · 1 mark · MCQBoard level
The 6th term of the G.P. 2, −6, 18, … is:
(a) 486
(b) −486
(c) 1458
(d) −162
Hint
r = −3.
Answer
(b) $2 (- 3)^{5} = - 486$
Q3 · 1 mark · MCQHarder board level
The sum of the first 10 terms of the A.P. whose nth term is $2 n + 1$ is:
(a) 110
(b) 120
(c) 130
(d) 210
Hint
First term 3, 10th term 21.
Answer
(b) $5 (3 + 21) = 120$
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): $1 + 2 + 3 + \dots + 50 = 1275$ .
Reason (R): The sum of the first $n$ natural numbers is $\frac{n ( n + 1 )}{2}$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
50 × 51 ÷ 2.
Answer
(c)
Q5 · 3 marks · Short answerHarder board level
Which term of the A.P. 3, 10, 17, … is 84 more than its 13th term?
Hint
Find T₁₃ first.
Answer
$T_{13} = 3 + 12 \times 7 = 87$ ; required term $= 171 = 3 + (n - 1) 7 \Rightarrow n = 25$ .
Q6 · 4 marks · Short answerHarder board level
The 3rd term of a G.P. is 24 and the 6th term is 192. Find the sum of its first 8 terms.
Hint
Divide the two terms to get r³.
Answer
$\frac{a r ^{5}}{a r ^{2}} = r^{3} = 8$ , $r = 2$ , $a = 6$ . $S_{8} = 6 (2^{8} - 1) = 1530$ .
Q7 · 3 marks · Long answerBoard level
Find the sum of the first 15 terms of the A.P. 2, 5, 8, …
Hint
a = 2, d = 3.
Answer
$S_{15} = \frac{15}{2} [4 + 14 \times 3] = \frac{15}{2} \times 46 = 345$ .
Q8 · 4 marks · Long answerHarder board level
How many terms of the A.P. 24, 21, 18, … must be taken so that their sum is 78? Explain why there are two answers.
Hint
You get $n^{2} - 17 n + 52 = 0$ .
Answer
$\frac{n}{2} [48 - 3 (n - 1)] = 78 \Rightarrow n^{2} - 17 n + 52 = 0 \Rightarrow n = 4$ or $13$ . Terms 5 to 13 are 12, 9, 6, 3, 0, −3, −6, −9, −12, which add to 0, so the sum is 78 after 4 terms and again after 13 terms.
Q9 · 4 marks · Long answerHOTS / competency
A ball is dropped from a height of 81 m. Each time it hits the ground it rises to $\frac{2}{3}$ of the height it fell. Find the total distance it has travelled when it hits the ground for the 5th time.
Hint
Every rise is followed by an equal fall.
Answer
Down 81; then up and down 54, 36, 24, 16 (a G.P. with $r = \frac{2}{3}$ ). Total $= 81 + 2 (54 + 36 + 24 + 16) = 81 + 2 \times 130 = 341$ m. (G.P. sum: $\frac{54 ( 1 - ( 2/3 ) ^{4} )}{1/3} = 130$ .)
Q10 · 4 marks · Long answerHarder board level
If 3072 is the $n$ th term of the G.P. 3, 12, 48, …, find $n$ and the sum of the first $n$ terms.
Hint
4ⁿ⁻¹ = 1024.
Answer
$3 \times 4^{n - 1} = 3072 \Rightarrow 4^{n - 1} = 1024 = 4^{5} \Rightarrow n = 6$ . $S_{6} = \frac{3 ( 4 ^{6} - 1 )}{3} = 4095$ .
Q11 · 1 mark · MCQHOTS / competency
Which of these A.P.s has 100 as a term?
(a) 3, 7, 11, …
(b) 4, 9, 14, …
(c) 1, 4, 7, …
(d) 2, 8, 14, …
Hint
(100 − a) must be a whole-number multiple of d.
Answer
(c): $100 - 1 = 99 = 33 \times 3$ , so 100 is the 34th term. The others give 24.25, 19.2 and 16.33…
Q12 · 4 marks · Short answerHOTS / competency
Three numbers in G.P. have product 216 and sum 19. Find them.
Hint
Take them as a/r, a, ar.
Answer
$a^{3} = 216 \Rightarrow a = 6$ . $\frac{6}{r} + 6 + 6 r = 19 \Rightarrow 6 r^{2} - 13 r + 6 = 0 \Rightarrow r = \frac{3}{2}$ or $\frac{2}{3}$ . Numbers: 4, 6, 9.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the nth term of an A.P.Show answer
$T_{n} = a + (n - 1) d$
2.Write two forms of the A.P. sum.Show answer
$\frac{n}{2} [2 a + (n - 1) d]$ and $\frac{n}{2} (a + l)$
3.Write the nth term of a G.P.Show answer
$T_{n} = a r^{n - 1}$
4.Write the G.P. sum for r > 1.Show answer
$\frac{a ( r ^{n} - 1 )}{r - 1}$
5.How do you find a term from a sum formula?Show answer
$T_{n} = S_{n} - S_{n - 1}$
6.How do you write three numbers in A.P. to make the sum simple?Show answer
a − d, a, a + d
7.What is r for 64, 32, 16, …?Show answer
½
8.Is 2, 4, 8, 14 an A.P., a G.P. or neither?Show answer
Neither: differences 2, 4, 6; ratios 2, 2, 1.75.

Best order to study it

A.P.: d, general term, terms from two given terms.
A.P. sums, including finding n (quadratic) and Tₙ = Sₙ − Sₙ₋₁.
G.P.: r, general term, which term equals a given number (powers).
G.P. sums for r > 1 and r < 1.
Applications and three-term problems; mixed A.P./G.P. questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you tell an A.P. from a G.P. and find d or r?
Do you always use (n − 1) in the general term?
Can you find a and d from two given terms?
Can you use Sₙ = n/2(a + l) when the last term is known?
Can you find n from a sum, and decide which roots are valid?
Can you find which term of a G.P. equals a given number using powers?
Can you set up an application as an A.P. or G.P. and decide term or sum?

Arithmetic and Geometric Progression: frequently asked questions

Is the sum of an infinite G.P. in the ICSE Class 10 syllabus?

The 2027 syllabus says 'finding sum of their first n terms', so practise sums of a finite number of terms.

Which sum formula should I use for an A.P.?

If you know the last term, use n/2(a + l). Otherwise use n/2[2a + (n − 1)d]. Both give the same answer.

Can the common difference or ratio be negative?

Yes. A negative d gives a decreasing A.P.; a negative r gives a G.P. whose signs alternate, like 2, −6, 18, …

Why can two values of n give the same A.P. sum?

If the A.P. goes from positive to negative terms, the extra negative terms can cancel some positive ones, so the running total returns to the same value. Both values are valid unless the question rules one out.

How do I find n when the G.P. term is a big number?

Divide by a, then write the result as a power of r (for example 729 = 3⁶). Knowing powers of 2, 3, 4 and 5 makes this quick.

Based on: CISCE ICSE 2027 Mathematics syllabus; CISCE ICSE 2027 Mathematics specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2025; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2024