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