Algebra - Geometric Progressions

The sum of the first three terms of a geometric progression (G.P) is $31$ and the sum of the first and the third term is $26$. Find the seventh term of the progression.

Let the given G.P be $b_1, \; b_2, \; b_3, \; b_4, \cdots$

Let the first term of the G.P be $= b_1 = \dfrac{a}{r}$;

second term $= b_2 = a$

and the third term $= b_3 = a \cdot r$

where $r$ is the common ratio of the G.P

$n^{th}$ term of the progression $= b_n = b_1 \times r^{n - 1}$

$\therefore \;$ Seventh term of G.P $= b_7 = b_1 \times r^6 = \dfrac{a}{r} \times r^6 = a\cdot r^5$

Given: $\;$ $b_1 + b_2 + b_3 = 31$

i.e. $\;$ $\dfrac{a}{r} + a + a \cdot r = 31$

i.e. $\;$ $\dfrac{a}{r} + a \cdot r = 31 - a$ $\;\;\; \cdots \; (1)$

And: $\;$ $b_1 + b_3 = 26$

i.e. $\;$ $\dfrac{a}{r} + a \cdot r = 26$ $\;\;\; \cdots \; (2)$

i.e. $\;$ $31 - a = 26$ $\;\;\;$ [in view of equation $(1)$]

$\implies$ $a = 5$

Substituting $a = 5$ in equation $(2)$ gives

$5 \left(\dfrac{1}{r} + r\right) = 26$

i.e. $\;$ $5 r^2 - 26 r + 5 = 0$

i.e. $\;$ $\left(5r - 1\right) \left(r - 5\right) = 0$

$\implies$ $r = \dfrac{1}{5}$ $\;$ or $\;$ $r = 5$

$\therefore \;$ When $a = 5$, $r = \dfrac{1}{5}$, the $7^{th}$ term of G.P $= 5 \times \left(\dfrac{1}{5}\right)^5 = \dfrac{1}{625}$

and when $a = 5$, $r = 5$, the $7^{th}$ term of G.P $= 5 \times 5^5 = 15625$

Algebra - Geometric Progressions

The sum of three numbers which form a geometric progression (G.P) is $13$ and the sum of their squares is $91$. Find the numbers.

Let the three numbers in G.P be $\;$ $\dfrac{a}{r}, \; a, \; a \cdot r$

where $r$ is the common ratio of the G.P

Given: $\;$ $\dfrac{a}{r} + a + a \cdot r = 13$

i.e. $\;$ $a \times \left(\dfrac{1}{r} + 1 + r\right) = 13$ $\;\;\; \cdots \; (1a)$

i.e. $\;$ $\dfrac{a}{r} \times \left(1 + r + r^2\right) = 13$

i.e. $\;$ $\dfrac{a^2}{r^2} = \dfrac{169}{\left(1 + r + r^2\right)^2}$ $\;\;\; \cdots \; (1b)$

And: $\;$ $\left(\dfrac{a}{r}\right)^2 + a^2 + \left(a \cdot r\right)^2 = 91$

i.e. $\;$ $a^2 \times \left(\dfrac{1}{r^2} + 1 + r^2\right) = 91$

i.e. $\;$ $\dfrac{a^2}{r^2} \times \left(1 + r^2 + r^4\right) = 91$

i.e. $\;$ $\dfrac{169}{\left(1 + r + r^2\right)^2} \times \left(1 + r^2 + r^4\right) = 91$ $\;\;\;$ [by equation $(1b)$]

i.e. $\;$ $\left(\dfrac{169}{1 + r + r^2}\right) \times \left(\dfrac{1 + r^2 + r^4}{1 + r + r^2}\right) = 91$

i.e. $\;$ $\left(\dfrac{169}{1 + r + r^2}\right) \times \left[r^2 + \dfrac{1 - r^3}{1 + r + r^2}\right] = 91$

i.e. $\;$ $\left(\dfrac{169}{1 + r + r^2}\right) \times \left[r^2 + \dfrac{\left(1 - r\right) \left(1 + r + r^2\right)}{1 + r + r^2}\right] = 91$

i.e. $\;$ $\left(\dfrac{169}{1 + r + r^2}\right) \times \left(r^2 - r + 1\right) = 91$

i.e. $\;$ $169 r^2 - 169 r + 169 = 91 + 91 r + 91 r^2$

i.e. $\;$ $78 r^2 - 260 r + 78 = 0$

i.e. $\;$ $3 r^2 - 10 r + 3 = 0$

i.e. $\;$ $\left(r - 3\right) \left(3r - 1\right) = 0$

i.e. $\;$ $r = 3$ $\;$ or $\;$ $r = \dfrac{1}{3}$

Substituting $r = 3$ in equation $(1a)$ gives

$a \times \left(\dfrac{1}{3} + 1 + 3\right) = 13$

i.e. $\;$ $a \times \dfrac{13}{3} = 13$ $\implies$ $a = 3$

Substituting $r = \dfrac{1}{3}$ in equation $(1a)$ gives

$a \times \left(\dfrac{1}{\dfrac{1}{3}} + 1 + \dfrac{1}{3}\right) = 13$

i.e. $\;$ $a \times \left(3 + 1 + \dfrac{1}{3}\right) = 13$

i.e. $\;$ $a \times \dfrac{13}{3} = 13$ $\implies$ $a = 3$

$\therefore \;$ When $a = 3$, $\;$ $r = 3$, the numbers are:

$\dfrac{3}{3}, \; 3 , \; 3 \times 3$ $\;$ i.e. $\;$ $1, \; 3, \; 9$

When $a = 3$, $\;$ $r = \dfrac{1}{3}$, the numbers are:

$\dfrac{3}{\dfrac{1}{3}}, \; 3 , \; 3 \times \dfrac{1}{3}$ $\;$ i.e. $\;$ $9, \; 3, \; 1$

$\implies$ the required numbers are $\;$ $1, \; 3, \; 9$

Algebra - Geometric Progressions

The first term of the geometric progression (G.P) $b_1, \; b_2, \; b_3, \cdots$ is unity. For what value of the common ratio of the progression is $4 b_2 + 5 b_3$ at a minimum?

First term of the G.P $= b_1 = 1$ $\;\;\;$ (Given)

Let the required common ratio $= r$

$n^{th}$ term of G.P $= b_n = b_1 \cdot r^{n - 1}$

$\therefore \;$ Second term of G.P $= b_2 = b_1 \cdot r = r$

Third term of G.P $= b_3 = b_1 \cdot r^2 = r^2$

Let $\;$ $S = 4 b_2 + 5 b_3$

i.e. $\;$ $S = 4 r + 5 r^2$

For $S$ to be a minimum, $\dfrac{dS}{dr} = 0$

i.e. $\;$ $\dfrac{dS}{dr} = 4 + 10 r = 0$

i.e. $\;$ $r = \dfrac{-4}{10} = \dfrac{-2}{5}$

$\therefore \;$ $S$ is a minimum when $r = \dfrac{-2}{5}$

Algebra - Geometric Progressions

The sum of the first and the third term of a geometric progression (G.P) is $20$ and the sum of the first three terms is $26$. Find the progression.

Let the first term of the G.P $= t_1 = a$ $\;$ and the common ratio $= r$

$n^{th}$ term of G.P $= t_n = a \cdot r^{n - 1}$

$\therefore \;$ Second term of G.P $= t_2 = a \cdot r$

Third term of G.P $= t_3 = a \cdot r^2$

Given: $\;$ $t_1 + t_3 = 20$

i.e. $\;$ $a + a \cdot r^2 = 20$

i.e. $\;$ $a \left(1 + r^2\right) = 20$ $\implies$ $a = \dfrac{20}{1 + r^2}$ $\;\;\; \cdots \; (1)$

And: $\;$ $t_1 + t_2 + t_3 = 26$

i.e. $\;$ $a + a \cdot r + a \cdot r^2 = 26$

i.e. $\;$ $a \left(1 + r + r^2\right) = 26$

i.e. $\;$ $\left(\dfrac{20}{1 + r^2}\right) \times \left(1 + r + r^2\right) = 26$ $\;\;\;$ [In view of equation $(1)$]

i.e. $\;$ $\dfrac{1 + r + r^2}{1 + r^2} = \dfrac{26}{20} = \dfrac{13}{10}$

i.e. $\;$ $10 + 10 r + 10 r^2 = 13 + 13 r^2$

i.e. $\;$ $3 r^2 - 10 r + 3 = 0$

i.e. $\;$ $\left(r - 3\right) \left(3r - 1\right) = 0$

i.e. $\;$ $r = 3$ $\;$ or $\;$ $r = \dfrac{1}{3}$

When $r = 3$, we have from equation $(1)$, $a = \dfrac{20}{1 + 3^2} = 2$

When $r = \dfrac{1}{3}$, we have from equation $(1)$, $a = \dfrac{20}{1 + \left(\dfrac{1}{3}\right)^2} = 18$

$\therefore \;$ The required G.P is

when $a = 2, \; r = 3$: $\;\;\;$ $2, \; 6, \; 18, \; 54, \cdots$

when $a = 18, \; r = \dfrac{1}{3}$: $\;\;\;$ $18, \; 6, \; 2, \; \dfrac{2}{3}, \cdots$

Algebra - Geometric Progressions

The numbers $a, \; b, \; c \;$ and $\; d$ $\;$ form a geometric progression (G.P).
Find $\;$ $\left(a - c\right)^2 + \left(b - c\right)^2 + \left(b - d\right)^2 - \left(a - d\right)^2$

Since the numbers $a$, $b$, $c$ and $d$ form a G.P

$\implies$ $\dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c} = k$ $\;\;$ where $k$ is a constant

Now, $\;$ $\dfrac{b}{a} = k$ $\implies$ $b = ak$ $\;\;\; \cdots \; (1)$

$\dfrac{c}{b} = k$ $\implies$ $c = bk = ak^2$ $\;\;\;$ [by equation $(1)$] $\;\;\; \cdots \; (2)$

$\dfrac{d}{c} = k$ $\implies$ $d = ck = ak^3$ $\;\;\;$ [by equation $(2)$] $\;\;\; \cdots \; (3)$

Now, $\;$ $\left(a - c\right)^2 + \left(b - c\right)^2 + \left(b - d\right)^2 - \left(a - d\right)^2$

$= \left(a - ak^2\right)^2 + \left(ak - ak^2\right)^2 + \left(ak - ak^3\right)^2 - \left(a - ak^3\right)^2$
$\hspace{4cm}$ [by equations $(1)$, $(2)$ and $(3)$]

$= a^2 \left(1 - k^2\right)^2 + a^2 k^2 \left(1 - k\right)^2 + a^2 k^2 \left(1 - k^2\right)^2 - a^2 \left(1 - k^3\right)^2$

$= a^2 \left[\left(1 - k^2\right)^2 - \left(1 - k^3\right)^2\right] + a^2 k^2 \left[\left(1 - k\right)^2 + \left(1 - k^2\right)^2\right]$

$= a^2 \left[\left(1 + k\right)^2 \left(1 - k\right)^2 - \left(1 - k\right)^2 \left(1 + k + k^2\right)^2\right]$
$\hspace{3cm}$ $+ a^2 k^2 \left[\left(1 - k\right)^2 + \left(1 + k\right)^2 \left(1 - k\right)^2\right]$

$= a^2 \left(1 - k\right)^2 \left[\left(1 + k\right)^2 - \left(1 + k + k^2\right)^2\right] + a^2 k^2 \left(1 - k\right)^2 \left[1 + \left(1 + k\right)^2\right]$

$= a^2 \left(1 - k\right)^2 \left(1 + k + 1 + k + k^2\right) \left(1 + k - 1 - k - k^2\right)$
$\hspace{4cm}$ $+ a^2 k^2 \left(1 - k\right)^2 \left(1 + 1 + 2k + k^2\right)$

$= a^2 \left(1 - k\right)^2 \left(2 + 2k + k^2\right) \left(-k^2\right) + a^2 k^2 \left(1 - k\right)^2 \left(2 + 2k + k^2\right)$

$= 0$

Algebra - Geometric Progressions

Find the first term and the common ratio of a geometric progression if the sum of its first three terms is $10.5$ and the difference between the first and the fourth term is equal to $31.5$.

Let the first term of the G.P be $= t_1 = a$ $\;$ and common ratio $= r$

$n^{th}$ term of G.P $= t_n = a \cdot r^{n-1}$

$\therefore \;$ $2^{nd}$ term $= t_2 = a \cdot r$, $\;$ $3^{rd}$ term $= t_3 = a \cdot ar^2$, $\;$ $4^{th}$ term $= t_4 = a \cdot r^3$

Given: $\;$ $t_1 + t_2 + t_3 = 10.5$

i.e. $\;$ $a + a \cdot r + a \cdot r^2 = 10.5$

i.e. $\;$ $a \left(1 + r + r^2\right) = 10.5$ $\;\;\; \cdots \; (1)$

And: $\;$ $t_4 - t_1 = 31.5$

i.e. $\;$ $a \cdot r^3 - a = 31.5$

i.e. $\;$ $a \left(r^3 - 1\right) = 31.5$

i.e. $\;$ $a = \dfrac{31.5}{r^3 - 1}$ $\;\;\; \cdots \; (2)$

$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes

$\left[\dfrac{31.5}{r^3 - 1}\right] \times \left(1 + r + r^2\right) = 10.5$

i.e. $\;$ $\left[\dfrac{31.5}{\left(r - 1\right) \left(r^2 + r + 1\right)}\right] \times \left(1 + r + r^2\right) = 10.5$

i.e. $\;$ $\dfrac{31.5}{r - 1} = 10.5$

i.e. $\;$ $31.5 = 10.5 r - 10.5$

i.e. $\;$ $10.5 r = 42$ $\implies$ $r = \dfrac{42}{10.5} = 4$

Substituting the value of $r$ in equation $(2)$ gives

$a = \dfrac{31.5}{4^3 - 1} = \dfrac{31.5}{63} = 0.5$

$\therefore \;$ The first term of the given G.P is $= a = 0.5$ and its common ratio is $= r = 4$

Algebra - Geometric Progressions

The sum of the first three terms of an increasing geometric progression (G.P) is $13$ and their product is $27$. Calculate the sum of the first five terms of the progression.

Let the first term of the G.P be $= t_1 = \dfrac{a}{r}$,

the second term $= t_2 = a$ $\;$ and the third term $= t_3 = a \cdot r$

where $\;$ $r$ $\;$ is the common ratio of the G.P

Given: $\;$ $t_1 + t_2 + t_3 = 13$

i.e. $\;$ $\dfrac{a}{r} + a + a \cdot r = 13$

i.e. $\;$ $a \left(\dfrac{1}{r} + 1 + r\right) = 13$ $\;\;\; \cdots \; (1)$

And: $\;$ $t_1 \times t_2 \times t_3 = 27$

i.e. $\;$ $\dfrac{a}{r} \times a \times a \cdot r = 27 = 3^3$

i.e. $\;$ $a^3 = 3^3$ $\implies$ $a = 3$

Substituting $a = 3$ in equation $(1)$ gives

$3 \left(\dfrac{1}{r} + 1 + r\right) = 13$

i.e. $\;$ $3 + 3r + 3r^2 = 13r$

i.e. $\;$ $3r^2 - 10r + 3 = 0$

i.e. $\;$ $\left(3r - 1\right) \left(r - 3\right) = 0$

i.e. $\;$ $r = \dfrac{1}{3}$ $\;$ or $\;$ $r = 3$

But since the first three terms of the given G.P are in an increasing order

$\implies$ $r > 1$

$\therefore \;$ $r = \dfrac{1}{3}$ $\;$ is not an acceptable solution.

Now, sum of first $n$ terms of a G.P $= S_n = \dfrac{t_1 \left(r^n - 1\right)}{r - 1}$

i.e. $\;$ $S_n = \dfrac{\dfrac{a}{r} \left(r^n - 1\right)}{r - 1}$

$\therefore \;$ Sum of first $5$ terms of G.P

$= S_5 = \dfrac{\dfrac{3}{3} \left(3^5 - 1\right)}{3 - 1} = \dfrac{1 \times 242}{2} = 121$