Question 26 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Class X1 - Maths -Sequences and Series Page 193
Answers
Answered by
8
Let the two numbers be a and b.
We know that r = (b/a)^1/n+1
r = (81/3)^1/2+1
= (81/3)^1/3
= 3
a = a * r = 3 * 3 = 9
b = a * r * r = 3 * 3 * 3 = 27.
The two numbers are 9 and 27.
We know that r = (b/a)^1/n+1
r = (81/3)^1/2+1
= (81/3)^1/3
= 3
a = a * r = 3 * 3 = 9
b = a * r * r = 3 * 3 * 3 = 27.
The two numbers are 9 and 27.
Answered by
22
Let a, and b are two numbers .
in such a way that,
3 , a , b, 81 are in GP.
∵ Tn = arⁿ⁻¹
so,
T₁ = 3 , T₄ = 3r³ = 81 ⇒r = 3
T₂ = a = 3r = 3 × 3 = 9
T₃ = 3r² = 3 × 3² = 27
Hence , 9 and 27
in such a way that,
3 , a , b, 81 are in GP.
∵ Tn = arⁿ⁻¹
so,
T₁ = 3 , T₄ = 3r³ = 81 ⇒r = 3
T₂ = a = 3r = 3 × 3 = 9
T₃ = 3r² = 3 × 3² = 27
Hence , 9 and 27
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