Geometric Progression’s general term with “a” as the first term and “r” as the common ratio is defined by:

aₙ = a.rⁿ⁻¹

Where ‘n’ = the number of terms.

**Theorem:** To prove that G.P.’s nth term with “a” as the first term & “r” as the common ratio which is given by:

aₙ = a.rⁿ⁻¹

**Proof:**

Assuming, a₁,a₂,a₃,a₄,…, aₙ be the given G.P. then

a₁ = a ⇒ a.r¹⁻¹ = a.r⁰ = a

As ‘r’ represents the common ratio

∴ a₂ / a₁ = r ⇒ a₂ = a₁.r⇒ a₂ = a.r

a₃ / a₂ = r = a₃ = a₂ .r ⇒ a₃ = a.r² = a.r³⁻¹

a₄ / a₃ = r = a₄ = a₃.r ⇒ a₄ = (a.r²).r = a.r⁴⁻¹

Just continue in this way to get,

aₙ = a.rⁿ⁻¹

Hence, the sequence will be a, ar, ar²,….,arⁿ⁻¹ as it’s a finite or infinite term.

## Examples Regarding Geometric Progression’s General Term

**Illustration 1: Write the general term of G.P. if there’s s 5,10,20,40,… sequence present**

Solution : General term is given by – aₙ = a.rⁿ⁻¹ ————(1)

As per the given sequence

You can see that the first term is defined by “a” that is equal to 5; in short, a = 5 & r which is the common ratio, defined by “r” = 2

Therefore, Eq. (1) ⇒ aₙ = 5 * (2)ⁿ⁻¹

Solution = 5 * (2)ⁿ⁻¹

**Illustration 2: Write the general term of G.P. if the sequence is 1/8, -¼, ½, -1……**

Answer: The General term is given by aₙ = a.rⁿ⁻¹…………….(1)

As per the given sequence,

First term = a = ⅛; r = common ration = -2

Therefore, Eq. (1) ⇒ aₙ =⅛ ×(-2)ⁿ⁻¹

= (2)⁻³ . (-2)ⁿ⁻¹

= (2)⁻³ . (2)ⁿ⁻¹. (-1)ⁿ⁻¹

= (-1)ⁿ⁻¹ . (2)ⁿ⁻⁴

## Common Term of G.P.

Considering the sequence as – a, ar, ar², ar³,……

1st term = a

2nd term = ar

3rd term = ar²

Likewise, the nth term, tn = arn-1

Hence, the Common ratio = “r” = (Any term) / (the Preceding term)

= tn / tn-1

= (arⁿ⁻¹) /(arⁿ⁻²)

= r

Therefore, G.P’s general term is expressed as arⁿ⁻¹ & G.P’s general form is a + ar + ar² + …..

**Taking an example: **

r = t2 / t1 = ar / a = r