Question

if G is the first of n geometric means between a and b, show that G^n+1=a^n b​

Answer 1

Answer:

To prove:- G₁ⁿ⁺¹ = aⁿb

Step-by-step explanation:

Given that G1  is the first of n geometric means between a and b .Thus the series should be arranged in the following manner-

a , G₁ , G₂ , G₃ ...................., Gₙ , b

Clearly

Number of terms = n + 2

$b=ar^{(n+2)-1}\\\;\\b=ar^{n+1}\\\;\\r=(\frac{b}{a})^{\frac{1}{n+1}}$

Now,

$G_1=ar\\\;\\G_1=a\times(\frac{b}{a})^{\frac{1}{n+1}}\\\;\\\text{Taking (n+1)th power both sides}\\\;\\(G_1)^{n+1}=a^{n+1}\times(\frac{b}{a})^{\frac{n+1}{n+1}}\\\;\\(G_1)^{n+1}=a^{n+1}\times\frac{b}{a}\\\;\\(G_1)^{n+1}=a^n\times b\\\;\\(G_1)^{n+1}=a^nb$

Note:- The nth term of a G.P. having first term as 'a' and common ratio 'r' is given by arⁿ⁻¹.