UPSEE Solved Model Paper 6

We can solve this problem by using Geometric mean between two given numbers. Geometric mean between two given numbers : Let a and b be two given positive numbers and G be the Geometric Mean between them. Then a, G, b are in a G.P. Thus, $\frac{G}{a}$ = $\frac{b}{G}$ (or) $G^2$ = ab (or) G = $\sqrt{ab}$ (Since G > 0) Inserting n-Geometric means between two given numbers : Let $G_1, G_2$, …., $G_n$ be the n geometric means between two given numbers a and b. Then, a, $G_1, G_2$, …., $G_n$, b are in G.P Now, b = (n+2)th terms of G.P. = $ar^{n+1}$ ; where r is the common ratio b = $ar^{n+1}$ $r^{n+1}$ = $\frac{b}{a}$ (or) r = $\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ and $G_1$ = ar = $a\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ $G_2$ = $ar^2$ = $a\left(\frac{b}{a}\right)^{\frac{2}{n+1}}$ ... $G_n$ = $ar^n$ = $a\left(\frac{b}{a}\right)^{\frac{n}{n+1}}$ y using this definition we can find the two geometric means between 1 and 64. The two geometric means between 1 and 64. 1, $G_1, G_2$, 64 are in G.P b = $ar^{n+1}$ ; here , n = 2 b = $ar^{2+1}$ = $ar^3$ ; here, a = 1 , b = 64 64 = $1 \times r^3$ $r^3$ = $4^3$ ∴ r = 4 $G_1$ = ar = 1 × 4 = 4 $G_2$ = $ar^2$ = 1 × $4^2$ = 16 Therefore, the two geometric means between 1 and 64 are 4 & 16.