Question

Q1.if G is the first of n geometric means between a and b, show that G^n+1=a^n b​

Answer 1

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

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

Now,

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

Answer 2

Step-by-step explanation:

Answer:

T o prove:- G₁ⁿ⁺¹ = aⁿb

Given tha t G1 is the first of n geometric means be tween 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

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

Now,

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

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