Selection of terms in geometric progression
selection of terms in geometric progression : Sometimes it is required to select the finite number of terms in G.P. It is always easy if we select the terms in the following manner.No. of terms  Terms  common ratio 

$\frac{a}{r}$, a,ar  

$\frac{a}{r^{3}}, \frac{a}{r},ar, ar^{3}$  

$\frac{a}{r^{2}}, \frac{a}{r},a,ar, ar^{2}$  
If the product of the numbers is not given, then the numbers are taken as
a, ar,a$r^{2}, ar^{3}$,...
Solution : Let the four numbers in G.P. be $\frac{a}{r^{3}}, \frac{a}{r}, ar, ar^{3}$
Product = 4096
$\frac{a}{r^{3}} \times \frac{a}{r} \times ar\times ar^{3}$ = 4096
∴ $a^{4}$ =4096
$a^{4} = 8^{4}$
∴ a = 8
$\frac{a}{r^{3}} + \frac{a}{r} + ar + ar^{3}$ = 8
a$\left (\frac{1}{r^{3}} + \frac{1}{r} + r + r^{3} \right )$ = 85
8$\left (\frac{1}{r^{3}} + \frac{1}{r} + r + r^{3} \right )$ = 85
8$\left ( r^{3} + \frac{1}{r^{3}} \right ) + 8\left ( r + \frac{1}{r} \right )$ = 85
8$\left [ \left ( r + \frac{1}{r} \right )^{3} 3\left(r +\frac{1}{r} \right) \right ]+ 8\left(r + \frac{1}{r} \right)$ = 85
8$\left ( r + \frac{1}{r} \right )^3  16\left ( r + \frac{1}{r} \right )$  85=0
8$x^{3}$  16x  85 = 0 Let x = r + $\frac{1}{r}$
Now factorize this quadratic equation
(2x 5)(4$x^{2}$ + 10 x + 17) = 0
⇒ 2x  5 = 0  since(4$x^{2}$ + 10 x + 17) has imaginary roots
∴ x = $\frac{5}{2}$
⇒ r + $\frac{1}{r} = \frac{5}{2}$ ⇒ 2$r^{2}$  5r + 2 = 0
So r = 2 or r = $\frac{1}{2}$
Put a = 8 and r = 2 or $\frac{1}{2}$ ,we get the four numbers are either 1,4,16,64 or 64,16,4,1
2) Find the three numbers in G.P whose sum is 52 and the sum of whose products in pairs is 624.
Solution : Let the required numbers be a, ar,a$r^{2}$
Sum = 52
⇒ a + ar + a$r^{2}$ = 52
⇒ a( 1 + r + $r^{2}$)= 52 (1) Sum of products in pairs = 624
i.e. a $\times$ ar + ar $\times ar^{2}$ + a $\times ar^{2}$ = 624
⇒ $a^{2}r (1 + r + r^{2}$) = 624 (2)
Divide equation (2) by (1) we get,
ar = 12 ⇒ a = $\frac{12}{r}$
Put a = $\frac{12}{r}$ in equation (1) we get
$\frac{12}{r}(( 1 + r + r^{2}$)= 52
⇒ 3$r^{2}$  10r + 3 = 0
Factorizing
(3r 1)(r 3) = 0
∴ r = $\frac{1}{3}$ or r = 3
If r = 3 then a = 4 and if r = $\frac{1}{3}$ then a = 36
∴ three terms of G.P are 4,12,36 or 36,12,4.
11th grade math
From selection of terms in geometric progression to Home
Examples on selection of terms in geometric progression
1) Find the four numbers in G.P whose sum is 85 and product is 4096.Solution : Let the four numbers in G.P. be $\frac{a}{r^{3}}, \frac{a}{r}, ar, ar^{3}$
Product = 4096
$\frac{a}{r^{3}} \times \frac{a}{r} \times ar\times ar^{3}$ = 4096
∴ $a^{4}$ =4096
$a^{4} = 8^{4}$
∴ a = 8
$\frac{a}{r^{3}} + \frac{a}{r} + ar + ar^{3}$ = 8
a$\left (\frac{1}{r^{3}} + \frac{1}{r} + r + r^{3} \right )$ = 85
8$\left (\frac{1}{r^{3}} + \frac{1}{r} + r + r^{3} \right )$ = 85
8$\left ( r^{3} + \frac{1}{r^{3}} \right ) + 8\left ( r + \frac{1}{r} \right )$ = 85
8$\left [ \left ( r + \frac{1}{r} \right )^{3} 3\left(r +\frac{1}{r} \right) \right ]+ 8\left(r + \frac{1}{r} \right)$ = 85
8$\left ( r + \frac{1}{r} \right )^3  16\left ( r + \frac{1}{r} \right )$  85=0
8$x^{3}$  16x  85 = 0 Let x = r + $\frac{1}{r}$
Now factorize this quadratic equation
(2x 5)(4$x^{2}$ + 10 x + 17) = 0
⇒ 2x  5 = 0  since(4$x^{2}$ + 10 x + 17) has imaginary roots
∴ x = $\frac{5}{2}$
⇒ r + $\frac{1}{r} = \frac{5}{2}$ ⇒ 2$r^{2}$  5r + 2 = 0
So r = 2 or r = $\frac{1}{2}$
Put a = 8 and r = 2 or $\frac{1}{2}$ ,we get the four numbers are either 1,4,16,64 or 64,16,4,1
2) Find the three numbers in G.P whose sum is 52 and the sum of whose products in pairs is 624.
Solution : Let the required numbers be a, ar,a$r^{2}$
Sum = 52
⇒ a + ar + a$r^{2}$ = 52
⇒ a( 1 + r + $r^{2}$)= 52 (1) Sum of products in pairs = 624
i.e. a $\times$ ar + ar $\times ar^{2}$ + a $\times ar^{2}$ = 624
⇒ $a^{2}r (1 + r + r^{2}$) = 624 (2)
Divide equation (2) by (1) we get,
ar = 12 ⇒ a = $\frac{12}{r}$
Put a = $\frac{12}{r}$ in equation (1) we get
$\frac{12}{r}(( 1 + r + r^{2}$)= 52
⇒ 3$r^{2}$  10r + 3 = 0
Factorizing
(3r 1)(r 3) = 0
∴ r = $\frac{1}{3}$ or r = 3
If r = 3 then a = 4 and if r = $\frac{1}{3}$ then a = 36
∴ three terms of G.P are 4,12,36 or 36,12,4.
From selection of terms in geometric progression to Home
Covid19 has led the world to go through a phenomenal transition .
Elearning is the future today.
Stay Home , Stay Safe and keep learning!!!
Covid19 has affected physical interactions between people.
Don't let it affect your learning.