# Geometric Progression problem

In this section we will discuss the geometric progression problem which are important according to the exam. But before seeing these problems, student should know all the formulas of geometric progression. If you are not conversant with the formulas then please learn the following formulas.## Formulas for geometric progression problem

If a1,a2,a3,....an are in G.P. then1) 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.

**Solution:**nth term of G.P is given by

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

∴ $a_{4} = ar^{4 - 1}$

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.

**Solution:**If A and G are arithmetic and geometric mean respectively between two positive numbers a and b, then the quadratic equation having a,b as its roots is $x^{2}$−2Ax + $G^{2}$ = 0

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.

**Solution:**Let a1,a2,a3,a4,a5 be 5 geometric means between a = 576 and b = 9

Sequence will be 576,

**a1, a2, a3, a4, a5,**9 is in G.P.

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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