1) nth term of G.P = $a_{n} = a r^{n - 1}$

2) sum of 'n' terms of a G.P = $S_{n} = a\left ( \frac{r^{n} - 1}{r - 1} \right )$ r > 1

For r $\neq $ 1,

$S_{n} = a\left ( \frac{1 - r^{n}}{1 - r} \right )$

3) $S_{n}$ = n and if r = l

$S_{n} = \left ( \frac{a - lr} {l - r} \right )$ OR $S_{n} = \left ( \frac{lr -a} {r - 1} \right )$ where 'l' is the last term.

4) G.M = $\sqrt{ab}$

5) A.M. =$\frac{a + b}{2}$

1)If the 4th,10th and 16th terms of a G.P. are x,y and z respectively. Prove that x,y,z are in G.P.

$a_{n} = ar^{n - 1}$

∴ $a_{4} = ar^{4 - 1}$

x = a$r^{3}$

$a_{10} = ar^{10 - 1}$

y = a$r^{9}$

$a_{16} = ar^{16 - 1}$

z = a$r^{15}$

x = a$r^{3}$ y = a$r^{9}$ z = a$r^{15}$

∴ $y^{2} = a^{2}r^{18}$ xy = a$r^{3}.ar^{15}$

xy = $a^{2}r^{18}= y^{2}$

∴ $y^{2}$= xy

Hence, x,y,z are in G.P.

2)If AM and GM of roots of quadratic equation are 8 and 5 respectively, then obtain the quadratic equation.

A = $\frac{a + b}{2}$ and $G^{2}=\sqrt{ab}$

8 = $\frac{a + b}{2}$

∴ (a + b) = 16

5 =$\sqrt{ab}$

25 = ab

The quadratic equation with roots a,b is $x^{2}$−(a+ b)x + ab = 0

$x^{2}$−16x + 25 = 0

3) Insert 5 geometric means between 576 and 9.

Sequence will be 576,

common ratio = r = $\left ( \frac{b}{a} \right )^{\frac{1}{n + 1}}$

r = $\left ( \frac{9}{576} \right )^{\frac{1}{5 + 1}}$ = $\left ( \frac{1}{64} \right )^{\frac{1}{6}} = \frac{1}{2}$

a1 = a$r^{1} = 576 \times \frac{1}{2}$ = 288

a2 = a$r^{2} = 576 \times \frac{1}{4}$ = 144

a3 = a$r^{3} = 576 \times \frac{1}{8}$ = 72

a4 = a$r^{4} = 576 \times \frac{1}{16}$ = 36

a5 = a$r^{5} = 576 \times \frac{1}{32}$ = 18

Hence 288,144,72,36,18 are the 5 geometric means inserted between 576 and 9.

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