Chapter 8: Arithmetic & Geometric Progressions

A sequence is an ordered list of numbers following a specific mathematical pattern. In this chapter, we will study two fundamental types of sequences: Arithmetic Progression (AP), where terms change by a constant addition/subtraction, and Geometric Progression (GP), where terms change by a constant multiplication/division.

Part A: Arithmetic Progression (A.P.)

1. What is an A.P.?

An Arithmetic Progression is a sequence of numbers in which the difference between any two consecutive terms is always constant. This constant is called the Common Difference ($d$).

Let the terms be $t_1, t_2, t_3, \ldots, t_n$.
For an AP: $t_2 - t_1 = t_3 - t_2 = t_4 - t_3 = \dots = d$.

Standard Form

If the first term is $a$ and the common difference is $d$, the AP is written as:

$$a, \quad a+d, \quad a+2d, \quad a+3d, \quad \dots$$

Example: $2, 5, 8, 11, \ldots$ is an AP with $a = 2$ and $d = 3$.

2. The $n^{\text{th}}$ Term (General Term) of an A.P.

To find any specific term in an AP without writing out the whole sequence, we use the formula for the $n^{\text{th}}$ term, denoted as $t_n$ or $a_n$.

$$t_n = a + (n - 1)d$$

Where:
$t_n$ = the $n^{\text{th}}$ term
$a$ = first term
$d$ = common difference
$n$ = position of the term (must be a positive integer: 1, 2, 3...)

Practice Problems — $n^{\text{th}}$ Term of A.P.

Q1. Find the 20th term of the A.P. $9, 13, 17, 21, \ldots$

Solution:
$a = 9$
$d = 13 - 9 = 4$
$t_n = a + (n-1)d$
$t_{20} = 9 + (20-1)(4) = 9 + 19(4) = 9 + 76 = 85$.
Ans: 85

Q2. Which term of the A.P. $3, 8, 13, 18, \ldots$ is $78$?

Solution:
$a = 3$, $d = 8 - 3 = 5$
Let $t_n = 78$
$a + (n-1)d = 78 \Rightarrow 3 + (n-1)5 = 78$
$5(n-1) = 75 \Rightarrow n-1 = 15 \Rightarrow n = 16$.
Ans: It is the 16th term.

Q3 (ICSE type). The 4th term of an A.P. is 22 and the 15th term is 66. Find the first term and the common difference.

Solution:
$t_4 = a + 3d = 22$ ...(i)
$t_{15} = a + 14d = 66$ ...(ii)
Subtract (i) from (ii): $11d = 44 \Rightarrow d = 4$.
Substitute $d$ in (i): $a + 3(4) = 22 \Rightarrow a + 12 = 22 \Rightarrow a = 10$.
Ans: a = 10, d = 4

3. Sum of First $n$ Terms of an A.P. ($S_n$)

To find the sum of the first $n$ terms ($S_n = t_1 + t_2 + \ldots + t_n$), we use either of the following formulas:

$$S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]$$

$$S_n = \frac{n}{2} (a + l)$$

(where $l$ is the last term, i.e., $l = t_n$)

Important Note

The relation between $S_n$ and $t_n$ is: $t_n = S_n - S_{n-1}$

Practice Problems — Sum of A.P.

Q1. Find the sum of the first 25 terms of the A.P. $5, 9, 13, 17, \ldots$

Solution:
$a = 5$, $d = 4$, $n = 25$
$S_{25} = \frac{25}{2}[2(5) + (25-1)(4)]$
$= \frac{25}{2}[10 + 24(4)] = \frac{25}{2}[10 + 96] = \frac{25}{2}[106] = 25 \times 53 = 1325$
Ans: 1325

Q2. Find the sum: $34 + 32 + 30 + \dots + 10$

Solution:
This is an A.P. with $a = 34$, $d = -2$, $l = 10$.
First, find $n$: $l = a + (n-1)d \Rightarrow 10 = 34 + (n-1)(-2)$
$-24 = -2(n-1) \Rightarrow 12 = n-1 \Rightarrow n = 13$.
Now sum: $S_n = \frac{n}{2}(a+l) = \frac{13}{2}(34 + 10) = \frac{13}{2}(44) = 13 \times 22 = 286$
Ans: 286

Q3. How many terms of the A.P. $24, 21, 18, \dots$ must be taken so that their sum is 78?

Solution:
$a = 24$, $d = -3$, $S_n = 78$
$S_n = \frac{n}{2}[2a + (n-1)d] \Rightarrow 78 = \frac{n}{2}[48 + (n-1)(-3)]$
$156 = n[48 - 3n + 3] = n(51 - 3n)$
$3n^2 - 51n + 156 = 0$. Divide by 3: $n^2 - 17n + 52 = 0$
$(n-4)(n-13) = 0 \Rightarrow n = 4$ or $n = 13$.
Ans: 4 terms or 13 terms. (Double answer is possible because terms from 5th to 13th add up to 0).

4. Three or More Terms in an A.P.

When the sum of terms in an A.P. is given, choosing symmetric terms simplifies calculations greatly because the '$d$' terms cancel out:

Three terms: $(a - d), \quad a, \quad (a + d)$ (Common difference = $d$)
Four terms: $(a - 3d), \quad (a - d), \quad (a + d), \quad (a + 3d)$ (Common diff = $2d$)
Five terms: $(a - 2d), \quad (a - d), \quad a, \quad (a + d), \quad (a + 2d)$ (Common diff = $d$)

Example: Find 3 numbers in A.P. whose sum is 15 and product is 80.

Solution:
Let terms be $(a-d), a, (a+d)$.
Sum $= 3a = 15 \Rightarrow a = 5$.
Product $= (5-d)(5)(5+d) = 80 \Rightarrow 25(25 - d^2) = 400$ Wait, $(5-d)(5+d) = 25-d^2$.
$5(25-d^2) = 80 \Rightarrow 25-d^2 = 16 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3$.
If $d=3$, terms: $2, 5, 8$. If $d=-3$, terms: $8, 5, 2$.
Ans: The numbers are 2, 5, and 8.

Part B: Geometric Progression (G.P.)

1. What is a G.P.?

A Geometric Progression is a sequence in which the ratio of any term to its preceding term is always constant. This constant is called the Common Ratio ($r$).

For a GP: $\frac{t_2}{t_1} = \frac{t_3}{t_2} = \frac{t_4}{t_3} = \dots = r$.

Standard Form

If the first term is $a$ and the common ratio is $r$, the GP is written as:

$$a, \quad ar, \quad ar^2, \quad ar^3, \quad \dots$$

Example: $3, 6, 12, 24, \ldots$ is a GP with $a = 3$ and $r = 2$.

2. The $n^{\text{th}}$ Term of a G.P.

The formula for finding the $n^{\text{th}}$ term ($t_n$) of a geometric progression is:

$$t_n = a \cdot r^{n-1}$$

Practice Problems — $n^{\text{th}}$ Term of G.P.

Q1. Find the 7th term of the G.P. $2, 6, 18, 54, \dots$

Solution:
$a = 2$, $r = \frac{6}{2} = 3$
$t_n = a \cdot r^{n-1}$
$t_7 = 2 \cdot (3)^{7-1} = 2 \cdot 3^6 = 2 \cdot 729 = 1458$
Ans: 1458

Q2. Which term of the G.P. $2, 2\sqrt{2}, 4, \dots$ is 128?

Solution:
$a = 2$, $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$
$t_n = a r^{n-1} \Rightarrow 128 = 2 (\sqrt{2})^{n-1}$
$64 = (\sqrt{2})^{n-1} \Rightarrow 2^6 = (2^{1/2})^{n-1} \Rightarrow 2^6 = 2^{\frac{n-1}{2}}$
Comparing powers: $6 = \frac{n-1}{2} \Rightarrow n-1 = 12 \Rightarrow n = 13$
Ans: It is the 13th term.

Q3. The 3rd term of a G.P. is 24 and the 6th term is 192. Find the 10th term.

Solution:
$t_3 = ar^2 = 24$
$t_6 = ar^5 = 192$
Divide: $\frac{ar^5}{ar^2} = \frac{192}{24} \Rightarrow r^3 = 8 \Rightarrow r = 2$
Put $r=2$ in $ar^2 = 24 \Rightarrow a(4) = 24 \Rightarrow a = 6$
$t_{10} = ar^9 = 6(2^9) = 6(512) = 3072$
Ans: 3072

3. Sum of First $n$ Terms of a G.P. ($S_n$)

To find the sum of the first $n$ terms of a GP, the formula depends on whether the common ratio $r$ is greater than 1 or less than 1.

If $r > 1$:
$$S_n = \frac{a(r^n - 1)}{r - 1}$$
If $r < 1$:
$$S_n = \frac{a(1 - r^n)}{1 - r}$$
Note: If $r = 1$, the sequence is $a, a, a \dots$ and $S_n = na$.

Practice Problems — Sum of G.P.

Q1. Find the sum of the first 8 terms of the G.P. $1, 3, 9, 27, \ldots$

Solution:
$a = 1$, $r = 3$. Since $r > 1$:
$S_n = \frac{a(r^n - 1)}{r - 1} \Rightarrow S_8 = \frac{1(3^8 - 1)}{3 - 1}$
$3^8 = 6561$.
$S_8 = \frac{6561 - 1}{2} = \frac{6560}{2} = 3280$
Ans: 3280

Q2. How many terms of the G.P. $3, 3^2, 3^3, \ldots$ are needed to give the sum 120?

Solution:
$a = 3$, $r = 3$, $S_n = 120$
$\frac{3(3^n - 1)}{3-1} = 120 \Rightarrow \frac{3(3^n - 1)}{2} = 120$
$3(3^n - 1) = 240 \Rightarrow 3^n - 1 = 80 \Rightarrow 3^n = 81$
$3^n = 3^4 \Rightarrow n = 4$
Ans: 4 terms.

4. Three or More Terms in a G.P.

When the product of terms in a G.P. is given, choose terms symmetrically so the '$r$' cancels when multiplied:

Three terms: $\frac{a}{r}, \quad a, \quad ar$ (Common ratio = $r$)
Four terms: $\frac{a}{r^3}, \quad \frac{a}{r}, \quad ar, \quad ar^3$ (Common ratio = $r^2$)

Summary Table

Property	Arithmetic Progression (A.P.)	Geometric Progression (G.P.)
Sequence Pattern	$a, a+d, a+2d \dots$	$a, ar, ar^2 \dots$
Key Parameter	Common Difference $d = t_n - t_{n-1}$	Common Ratio $r = \frac{t_n}{t_{n-1}}$
$n^{\text{th}}$ Term ($t_n$)	$a + (n-1)d$	$ar^{n-1}$
Sum of $n$ Terms ($S_n$)	$\frac{n}{2}[2a+(n-1)d]$ or $\frac{n}{2}(a+l)$	$a\frac{r^n-1}{r-1}$ if $r>1$, or $a\frac{1-r^n}{1-r}$ if $r<1$
Condition for 3 terms (A, B, C)	$2B = A + C$	$B^2 = A \times C$

Tips for ICSE Board Exams

Always state $a$, $d$ or $r$ explicitly before writing formulas.
In AP sum problems, beware of quadratic equations for $n$. A negative $n$ or fraction $n$ must be rejected.
In "Which term of the AP..." questions, let $t_n$ be that value and solve for $n$.
If $S_n$ is given as a quadratic like $3n^2 + 2n$, you can find $t_n$ by using $t_n = S_n - S_{n-1}$.
Do not confuse formulas for AP and GP. Check whether the problem says AP or GP before proceeding!