Question

if a,b,c are in A.P. then prove that 3^a,3^b,3^c are in G.P.

Answer 1

proved.....

i hope u understand the method i have done.

Answer 2

Answer:

It is proved that $\bold{3^{a}, 3^{b}, 3^{c}}$ is in G.P.

Step-by-step explanation:

It is given that a, b, c are in A.P, hence $\mathbf{b}=\frac{a+b}{2} [Since, a-b = b-c]$

For $3^{a}, 3^{b}, 3^{c}$ to be in G.P., $3^{b}$ has to be equal to $\sqrt{3^{a} \times 3^{c}}$ [Since, for x, y, z in G.P =>  y = √xz]

∴ $3^{b}=\sqrt{3^{a} \times 3^{c}}$

Squaring on both sides:

$=>\left(3^{b}\right)^{2}=\left(\sqrt{3}^{a} \times 3^{c}\right)^{2}$

On squaring the square root will get eliminated

$\begin{array}{l}{=>3^{2 b}=3^{a} \times 3^{c}} \\ {=>3^{2 b}=3^{a+c}} \\ {=>2 b=a+c}\end{array}$ [Since, bases are common.]

$=>b=\frac{a+c}{2} x$

Thus, it is proved that if a, b, c are in A.P., then $\bold{3^{a}, 3^{b}, 3^{c}}$ are in G.P.