Show that the progression 81 , 27 , 9 , 3 , 1 ... is a G.P. Write its (i) first term (ii) common ratio (iii) nth term (iv) 11th term.
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Answers
Answer :-
The given progression is 81, 27, 9, 3, 1, ...
We have, 27/81 = 9/27 = 3/9 = 1/3 [Constant]
So, the given progression is a G.P.
We have,
(i) First term, a = 81.
(ii) common ratio, r = 27/81 = 1/3.
(iii) nth term, Tₙ = arⁿ⁻¹
=> Tₙ = 81 × (1/3)ⁿ⁻¹ = 81/3ⁿ⁻¹ - (X)
(iv) Putting n = 11 in (X), we get:
=> T₁₁ = 81/(3¹¹ - 1) = 81/3¹⁰ = 3⁴/¹⁰ = 1/3⁶ = 1/729.
EXPLANATION.
Geometric progression.
⇒ 81, 27, 9, 3, 1 . . . . .
As we know that,
First term = a = 81.
Common ratio = b/a = 27/81 = 1/3.
As we know that,
General term of G.P.
⇒ Tₙ = a x rⁿ⁻¹.
Put the value in the equation, we get.
⇒ Tₙ = 81 x (1/3)ⁿ⁻¹.
11th term of G.P.
⇒ T₁₁ = ar¹⁰.
Put the values in the equation, we get.
⇒ T₁₁ = 81 x (1/3)¹⁰.
⇒ T₁₁ = 81 x (1/3) x (1/3) x (1/3) x (1/3) x (1/3)⁶.
⇒ T₁₁ = [1/(3)]⁶.
⇒ T₁₁ = 1/729.
MORE INFORMATION.
Geometric mean.
⇒ a, G, b are in G.P.
⇒ G² = ab.
⇒ G = ± √ab.
⇒ G = + √ab. [For positive numbers].
⇒ G = - √ab. [For negative numbers].
Nth Geometric mean.
⇒ a, G₁, G₂, G₃, . . . . . Gₙ , b. are in G.P.
⇒ T₁ = a.
⇒ Tₙ₊₂ = b.
⇒ Tₙ₊₂ = a. r ⁽ⁿ⁺²⁻¹⁾.
⇒ b = a x rⁿ⁺¹.
⇒ r⁽ⁿ⁺¹⁾ = b/a.
⇒ r = (b/a)^(1/n + 1)