# CLASS-10GEOMETRIC PROGRESSION - GENERAL TERM OF G.P.

General Term of G.P.-

Theorem.1- with 1st term ‘a’, common ratio ‘r’, show that its e-th term is given by Tₑ = arᵉˉ¹

Proof Let us consider a G.P. with 1st term = a, and common ratio = r.

Then, the G.P. is a, ar, ar², ar³, ar⁴, ar⁵,………………….

In this G.P. we have

1st term T₁ = a = ar⁰ = ar⁽¹ˉ¹⁾

2nd term T₂ = ar = ar¹ = ar⁽²ˉ¹⁾

3rd term T₃ = ar² = ar⁽³ˉ¹⁾

4th term T₄ = ar³ = ar⁽⁴ˉ¹⁾

……………. ……………. ………………

……………. ……………. ………………

So, e-th term, Tₑ = ar⁽ᵉˉ¹⁾

Hence, its e-th term is given by Tₑ = ar⁽ᵉˉ¹⁾

Example.1) Show that the progression 3, 6, 12, 24, 48,…………… is a G.P. write its

(i) first term (ii) common ratio

(iii) e-th term (iv) 12th term

Ans.) The given terms are 3, 6, 12, 24, 48,……………

6         12         24          48

Clearly, we have ------ = ------ = ------- = ------- = 2 (constant)

3          6          12          24

So, the given number form a G.P.

Clearly, we have

(i) first term a = 3        (Ans.)

6

(ii) common ratio, r = ------ = 2       (Ans.)

3

(iii) e-th term, Tₑ = arᵉˉ¹

=> Tₑ = (3 X 2ᵉˉ¹) ……………….(Ans.)

(iv) Putting e = 12, in Tₑ = arᵉˉ¹ where a = 3, r = 2 and we get –

Tₑ = arᵉˉ¹

=> T₁₂ = (3 X 2¹²ˉ¹)

=> (3 X 2¹¹)

=> (3 X 2048)

=>  6144 ……………(Ans.)