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.)
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