Arithmetic & Geometric Progression

ICSE Class 10 Mathematics • Chapter 08

PART A: Arithmetic Progression (AP)

Arithmetic Progression: A sequence of numbers where the difference between consecutive terms is constant.

Common Difference (d): $d = a_2 - a_1 = a_3 - a_2 = ...$

Example: 2, 5, 8, 11, ... (d = 3)

1. nth Term of AP

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

Where: $a$ = first term, $d$ = common difference, $n$ = position

2. Sum of n Terms of AP

Form 1 (when d is known):

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

Form 2 (when last term l is known):

$S_n = \frac{n}{2}[a + l]$

Example 1: Find 15th term and sum of first 15 terms of AP: 3, 7, 11, 15, ...

$a = 3$, $d = 7 - 3 = 4$

$a_{15} = 3 + (15-1) \times 4 = 3 + 56 = 59$

$S_{15} = \frac{15}{2}[3 + 59] = \frac{15}{2} \times 62 = 465$

3. Finding nth Term from Given Terms

Example 2: If 5th term is 19 and 8th term is 31, find AP.

$a_5 = a + 4d = 19$ ... (i)

$a_8 = a + 7d = 31$ ... (ii)

Subtracting (i) from (ii): $3d = 12$ → $d = 4$

From (i): $a + 16 = 19$ → $a = 3$

AP: 3, 7, 11, 15, ...

4. Properties of AP

If three terms are in AP: Choose $a-d$, $a$, $a+d$
If four terms are in AP: Choose $a-3d$, $a-d$, $a+d$, $a+3d$
Middle term = $\frac{a + l}{2}$ (average of first and last)

PART B: Geometric Progression (GP)

Geometric Progression: A sequence where the ratio between consecutive terms is constant.

Common Ratio (r): $r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = ...$

Example: 2, 6, 18, 54, ... (r = 3)

1. nth Term of GP

$a_n = ar^{n-1}$

Where: $a$ = first term, $r$ = common ratio, $n$ = position

2. Sum of n Terms of GP

When r ≠ 1:

$S_n = \frac{a(r^n - 1)}{r - 1}$ when $r > 1$

$S_n = \frac{a(1 - r^n)}{1 - r}$ when $r < 1$

When r = 1: $S_n = na$

Example 3: Find 6th term and sum of first 6 terms of GP: 2, 6, 18, ...

$a = 2$, $r = \frac{6}{2} = 3$

$a_6 = 2 \times 3^5 = 2 \times 243 = 486$

$S_6 = \frac{2(3^6 - 1)}{3 - 1} = \frac{2(729 - 1)}{2} = 728$

3. Properties of GP

If three terms are in GP: Choose $\frac{a}{r}$, $a$, $ar$
Geometric Mean of a and b: $\sqrt{ab}$
In GP: $b^2 = ac$ (for consecutive terms a, b, c)

Comparison: AP vs GP

Property	AP	GP
Definition	Constant difference	Constant ratio
Test	$a_2 - a_1 = a_3 - a_2$	$\frac{a_2}{a_1} = \frac{a_3}{a_2}$
nth term	$a + (n-1)d$	$ar^{n-1}$
Sum formula	$\frac{n}{2}[2a + (n-1)d]$	$\frac{a(r^n - 1)}{r - 1}$
Middle term	Arithmetic mean = $\frac{a+c}{2}$	Geometric mean = $\sqrt{ac}$

Quick Reference

Formula	AP	GP
nth term	$a_n = a + (n-1)d$	$a_n = ar^{n-1}$
Sum (n terms)	$\frac{n}{2}[2a + (n-1)d]$	$\frac{a(r^n - 1)}{r - 1}$
Sum (with last)	$\frac{n}{2}[a + l]$	-
To find n	$n = \frac{a_n - a}{d} + 1$	$n = \frac{\log(a_n/a)}{\log r} + 1$

Exam Practice (PYQ Trends)

PYQ: 2023

BOARD The 4th term of an AP is 11 and the 8th term exceeds the 4th term by 16. Find the AP and sum of first 20 terms.

PYQ: 2022

BOARD Find sum of integers between 100 and 300 that are divisible by 7.

Additional

HOTS First term of GP is 1. Sum of 3rd and 5th terms is 90. Find common ratio.