MathsISC Class 11 · Sequence and Series

Arithmetic and Geometric Progressions: nth Term, Sums, Means and Applications

Updated 17 October 2026 · Fact-checked

In an AP with first term a and common difference d: Tₙ = a + (n − 1)d and Sₙ = (n/2)[2a + (n − 1)d]. In a GP with first term a and ratio r: Tₙ = arⁿ⁻¹, Sₙ = a(rⁿ − 1)/(r − 1) and, for |r| < 1, S∞ = a/(1 − r). Means are inserted by finding d or r from the end terms.

In an AP every step adds the same d, so growth is steady; in a GP every step multiplies by r, so the bars grow faster and faster.

Illustration drawn for QuestionBuddy.

Understand Arithmetic and Geometric Progressions and Means

Arithmetic progression. $a, a + d, a + 2 d, ...$ with $T_{n} = a + (n - 1) d$ ; $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l)$ where l is the last term. Three numbers in AP are best taken as $a - d, a, a + d$ and four as $a - 3 d, a - d, a + d, a + 3 d$ . If $a, b, c$ are in AP, then $2 b = a + c$ .

Arithmetic means. To insert n AMs between a and b, the common difference is $d = \frac{b - a}{n + 1}$ . The sum of the n means equals n times the single AM $\frac{a + b}{2}$ .

Geometric progression. $a, a r, a r^{2}, ...$ with $T_{n} = a r^{n - 1}$ ; $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ for r ≠ 1 (equal to na if r = 1). Three numbers in GP: $\frac{a}{r}, a, a r$ . If $a, b, c$ are in GP then $b^{2} = a c$ . For $∣ r ∣ < 1$ , $S_{\infty} = \frac{a}{1 - r}$ .

Geometric means. To insert n GMs between a and b, $r = (\frac{b}{a})^{1/ (n + 1)}$ . Recurring decimals are infinite GPs: $0. \overline{27} = 0.27 + 0.0027 + ... = \frac{0.27}{1 - 0.01} = \frac{3}{11}$ .

Solving. Translate the data into equations in a, d (or a, r), solve simultaneously, then compute the required term or sum. Check with the first few terms.

Arithmetic and Geometric Progressions and Means: formulas to remember

AP: $T_n = a + (n - 1)d$ | $S_n = \frac{n}{2}[2a + (n - 1)d]$; Also $S_{n} = \frac{n}{2} (a + l)$ .
GP: $$T_n = ar^{n-1}$ | $S_n = \frac{a(r^n - 1)}{r - 1}$ | $S_∞ = \frac{a}{1 - r}$ (|r| < 1)$; r ≠ 1.
Inserting means: $AM: $d = \frac{b - a}{n + 1}$ | GM: $r = (\frac{b}{a})^{1/(n+1)}$$; n means between a and b.
Three terms: $AP: $a - d, a, a + d$ | GP: $\frac{a}{r}, a, ar$$; Simplifies the sum and product.

Arithmetic and Geometric Progressions and Means in real life

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.

A saving plan

Suppose a student saves Rs 5 in the first week and increases her saving by Rs 4 each week.

Find the saving in the 20th week and the total saved in 20 weeks.

Try it first, then show the solutionSolution

AP with $a = 5$ , $d = 4$ : $T_{20} = 5 + 19 \times 4 = 81$ .
$S_{20} = \frac{20}{2} (5 + 81) = 10 \times 86 = 860$ .
So she saves Rs 81 in the 20th week and Rs 860 in total.

Answer: Rs 81; Rs 860.

Memory hook: Use S = n/2 (first + last).

Tricks to remember Arithmetic and Geometric Progressions and Means

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

Symmetric choice
Take three AP terms as a − d, a, a + d so the sum is 3a.
Use it when: Sum and product problems.
Sum to infinity needs |r| < 1
Check the ratio before using a/(1 − r).
Use it when: Infinite GP.

Arithmetic and Geometric Progressions and Means: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A term of an AP · 1 mark · MCQBoard level

The 5th term of the AP 3, 7, 11, ... is:
(a) 15
(b) 17
(c) 19
(d) 23

Try it first, then show the solutionSolution

$a = 3$ , $d = 4$ : $T_{5} = 3 + 4 \times 4 = 19$ .

Answer: (c) 19

Example 2: Sum of an AP · 3 marks · Short answerBoard level

Find the sum of the first 20 terms of the AP 5, 9, 13, ...

Try it first, then show the solutionSolution

$a = 5$ , $d = 4$ , $n = 20$ . (0.5 mark)
$S_{20} = \frac{20}{2} [2 \times 5 + 19 \times 4]$ . (1.5 marks)
$= 10 \times 86 = 860$ . (1 mark)

Answer: 860

Example 3: AP from two terms · 4 marks · Short answerHarder board level

The 4th term of an AP is 14 and the 9th term is 29. Find the AP and the sum of its first 10 terms.

Try it first, then show the solutionSolution

$a + 3 d = 14$ and $a + 8 d = 29$ . (1 mark)
Subtract: $5 d = 15$ , $d = 3$ ; $a = 5$ . (1.5 marks)
$S_{10} = \frac{10}{2} [10 + 27] = 5 \times 37 = 185$ . (1.5 marks)

Answer: a = 5, d = 3; S₁₀ = 185.

Example 4: GP and means · 4 marks · Short answerHarder board level

The 3rd and 6th terms of a GP are 12 and 96. Find the GP and the sum of its first 8 terms. Insert 3 arithmetic means between 5 and 29 and 2 geometric means between 3 and 81.

Try it first, then show the solutionSolution

$a r^{2} = 12$ , $a r^{5} = 96$ gives $r^{3} = 8$ , $r = 2$ , $a = 3$ ; $S_{8} = 3 (2^{8} - 1) = 765$ . (2 marks)
AMs: $d = \frac{29 - 5}{4} = 6$ : 11, 17, 23. (1 mark)
GMs: $r^{3} = \frac{81}{3} = 27$ , $r = 3$ : 9 and 27. (1 mark)

Answer: GP 3, 6, 12, ... ; S₈ = 765; AMs 11, 17, 23; GMs 9, 27.

Example 5: Sum and product conditions; infinite GP · 5 marks · Long answerHOTS / competency

Three numbers in AP have sum 15 and product 105. Find them. Also find the sum to infinity of the GP 6, 2, 2/3, ... and express 0.272727... as a fraction.

Try it first, then show the solutionSolution

Take $a - d, a, a + d$ : $3 a = 15$ gives $a = 5$ . (1 mark)
Product: $5 (25 - d^{2}) = 105$ gives $d^{2} = 4$ , $d = \pm 2$ ; the numbers are 3, 5, 7. (1.5 marks)
GP: $a = 6$ , $r = \frac{1}{3}$ , $S_{\infty} = \frac{6}{1 - 1/3} = 9$ . (1.5 marks)
$0.272727... = \frac{0.27}{1 - 0.01} = \frac{27}{99} = \frac{3}{11}$ . (1 mark)

Answer: 3, 5, 7; 9; 3/11.

Where students lose marks in Arithmetic and Geometric Progressions and Means

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

Mistake: Using n instead of n − 1
Writing Tₙ = a + nd.
Do this: The nth term is a + (n − 1)d.
Usually costs: 2 marks.
Mistake: Wrong formula for the infinite GP
Using it when |r| ≥ 1.
Do this: The sum to infinity exists only for |r| < 1.
Usually costs: 1 to 2 marks.
Mistake: Number of means
Using d = (b − a)/n for n means.
Do this: There are n + 2 terms, n + 1 intervals: d = (b − a)/(n + 1).
Usually costs: 1 to 2 marks.

Exam tips

Write what is given and what is needed with a, d (or r).
Check your AP/GP by listing the first terms.
Use symmetric terms for sum-and-product problems.
State the condition |r| < 1 for infinite sums.

Practice questions on Arithmetic and Geometric Progressions and Means

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 sum of the first n natural numbers is:
(a) $n (n + 1)$
(b) $\frac{n ( n + 1 )}{2}$
(c) $\frac{n ^{2}}{2}$
(d) $n^{2}$
Hint
An AP with a = 1, d = 1.
Answer
(b) $\frac{n ( n + 1 )}{2}$
Q2 · 2 marks · Short answerBoard level
Find the 10th term and the sum of 10 terms of the GP 2, 6, 18, ...
Hint
a = 2, r = 3.
Answer
$T_{10} = 2 \times 3^{9} = 39366$ ; $S_{10} = 2 \times \frac{3 ^{10} - 1}{2} = 59048$ .
Q3 · 3 marks · Short answerHarder board level
How many terms of the AP 24, 21, 18, ... must be taken to give a sum of 78?
Hint
Quadratic in n.
Answer
$\frac{n}{2} [48 - 3 (n - 1)] = 78$ gives $n^{2} - 17 n + 52 = 0$ ; $n = 4$ or $n = 13$ (both valid as the terms turn negative after the 9th).
Q4 · 3 marks · Short answerHarder board level
If a, b, c are in AP and a², b², c² are in GP, show that the three are equal (when positive).
Hint
Use 2b = a + c and b⁴ = a²c².
Answer
$b = \frac{a + c}{2}$ and $b^{2} = a c$ give $\frac{( a + c ) ^{2}}{4} = a c$ , so $(a - c)^{2} = 0$ , $a = c$ , and hence $a = b = c$ .
Q5 · 3 marks · Short answerHarder board level
A ball is dropped from 80 m and rebounds to three-fifths of its previous height each time. Find the total distance travelled until it stops.
Hint
Down once, then up-and-down repeatedly.
Answer
Down: 80; each rebound goes up and down: $2 \times 80 \times \frac{3}{5}$ , then $r = \frac{3}{5}$ : total $80 + 2 \times 80 \times \frac{3/5}{1 - 3/5} = 80 + 160 \times 1.5 = 320$ m.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a loan is repaid in instalments that form an AP: the first is Rs 500 and each next one is Rs 50 more, for 12 months.
(i) Find the 12th instalment. (1 mark)
(ii) Find the total repaid. (1 mark)
(iii) Find the average instalment. (1 mark)
(iv) Find the month in which the instalment first exceeds Rs 800. (1 mark)
Hint
a = 500, d = 50.
Answer
(i) $500 + 11 \times 50 = 1050$ . (ii) $S_{12} = 6 (500 + 1050) = 9300$ . (iii) $9300/12 = 775$ . (iv) $500 + 50 (n - 1) > 800$ gives $n > 7$ , so the 8th month (Rs 850).

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.Tₙ of AP?Show answer
a + (n − 1)d.
2.Sₙ of GP?Show answer
a(rⁿ − 1)/(r − 1).
3.S∞?Show answer
a/(1 − r).
4.Condition for three numbers in GP?Show answer
b² = ac.

Arithmetic and Geometric Progressions and Means: frequently asked questions

What is the difference between a sequence and a series?

A sequence is a list of terms; a series is the sum of the terms of a sequence.

Can the common difference be negative?

Yes. Then the AP is decreasing.

Can r be negative in a GP?

Yes, giving terms that alternate in sign.