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

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

Algebra - Geometric Progressions

The sequence $\left\{b_n\right\}$ is a geometric progression (G.P) with $\dfrac{b_4}{b_6} = \dfrac{1}{4}$ and $b_2 + b_5 = 216$. Find $b_1$.

Let the first term of the G.P be $= b_1$

and the common ratio be $= r$

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

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

Fourth term of G.P $= b_4 = b_1 \cdot r^3$

Fifth term of G.P $= b_5 = b_1 \cdot r^4$

Sixth term of G.P $= b_6 = b_1 \cdot r^5$

Given: $\;$ $\dfrac{b_4}{b_6} = \dfrac{1}{4}$

i.e. $\;$ $\dfrac{b_1 \cdot r^3}{b_1 \cdot r^5} = \dfrac{1}{4}$

i.e. $\;$ $\dfrac{1}{r^2} = \dfrac{1}{4}$ $\implies$ $r^2 = 4$ $\implies$ $r = \pm 2$

And: $\;$ $b_2 + b_5 = 216$

i.e. $\;$ $b_1 \cdot r + b_1 \cdot r^4 = 216$

i.e. $\;$ $b_1 \left(r + r^4\right) = 216$ $\implies$ $b_1 = \dfrac{216}{r + r^4}$

$\therefore \;$ When $\;$ $r = + 2$, $\;$ $b_1 = \dfrac{216}{2 + 2^4} = \dfrac{216}{18} = 12$

When $\;$ $r = - 2$, $\;$ $b_1 = \dfrac{216}{- 2 + \left(-2\right)^4} = \dfrac{216}{14} = \dfrac{108}{7}$

Algebra - Geometric Progressions

Find three numbers which form a geometric progression (G.P) if their sum is $35$ and the sum of their squares is $525$.

Let the first term of the G.P be $= t_1 = a$

and the common ratio be $= r$

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

Given: $\;$ $\dfrac{a}{r} + a + a \cdot r = 35$

i.e. $\;$ $a \left(\dfrac{1}{r} + 1 + r\right) = 35$

$\implies$ $a = \dfrac{35 r}{1 + r + r^2}$ $\;\;\; \cdots \; (1)$

And: $\;$ $\left(\dfrac{a}{r}\right)^2 + a^2 + \left(a \cdot r\right)^2 = 525$

i.e. $\;$ $a^2 \times \left(\dfrac{1}{r^2} + 1 + r^2\right) = 525$ $\;\;\; \cdots \; (2)$

Substituting the value of $a$ from equation $(1)$ in equation $(2)$ gives

$\left(\dfrac{35 r}{1 + r + r^2}\right)^2 \times \left(\dfrac{1}{r^2} + 1 + r^2\right) = 525$

i.e. $\;$ $\dfrac{1 + r^2 + r^4}{\left(1 + r + r^2\right)^2} = \dfrac{525}{1225} = \dfrac{3}{7}$

i.e. $\;$ $\left(\dfrac{1}{1 + r + r^2}\right) \times \left(\dfrac{1 + r^2 + r^4}{1 + r + r^2}\right) = \dfrac{3}{7}$

i.e. $\;$ $\left(\dfrac{1}{1 + r + r^2}\right) \times \left[r^2 + \dfrac{1 - r^3}{1 + r + r^2}\right] = \dfrac{3}{7}$

i.e. $\;$ $\left(\dfrac{1}{1 + r + r^2}\right) \times \left[r^2 + \dfrac{\left(1 - r\right) \left(1 + r + r^2\right)}{1 + r + r^2}\right] = \dfrac{3}{7}$

i.e. $\;$ $\dfrac{r^2 + 1 - r}{1 + r + r^2} = \dfrac{3}{7}$

i.e. $\;$ $7 r^2 + 7 - 7 r = 3 + 3r + 3 r^2$

i.e. $\;$ $4 r^2 - 10r + 4 = 0$

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

i.e. $\;$ $\left(2r - 1\right) \left(r - 2\right) = 0$

i.e. $\;$ $r = \dfrac{1}{2}$ $\;$ or $\;$ $r = 2$

Substituting for $r$ in equation $(1)$ gives

when $\;$ $r = \dfrac{1}{2}$, $\;$ $a = \dfrac{35 \times \dfrac{1}{2}}{1 + \dfrac{1}{2} + \left(\dfrac{1}{2}\right)^2} = \dfrac{35 \times 4}{7 \times 2} = 10$

when $\;$ $r = 2$, $\;$ $a = \dfrac{35 \times 2}{1 + 2 + 2^2} = \dfrac{70}{7} = 10$

$\therefore \;$ The three numbers in G.P when $\;$ $a = 10$, $\;$ $r = \dfrac{1}{2}$ $\;$ are

$\dfrac{10}{1/2}, \; 10, \; 10 \times \dfrac{1}{2}$ $\;\;\;$ i.e. $\;$ $20, \; 10, \; 5$

or when $\;$ $a = 10$, $\;$ $r = 2$ $\;$ are

$\dfrac{10}{2}, \; 10, \; 10 \times 2$ $\;\;\;$ i.e. $\;$ $5, \; 10, \; 20$

$\implies$ The three numbers are $\;$ $5, \; 10, \; 20$

Algebra - Geometric Progressions

The sum of the first two terms of a geometric progression (G.P) is $15$. The first term exceeds the common ratio of the progression by $\dfrac{25}{3}$. Find the fourth term of the progression.

Let the first term of the G.P be $= t_1 = a$

and the common ratio be $= r$

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

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

Given: $\;$ $t_1 + t_2 = 15$

i.e. $\;$ $a + a \cdot r = 15$

i.e. $\;$ $a \left(1 + r\right) = 15$ $\;\;\; \cdots \; (1)$

And: $\;$ $t_1 = r + \dfrac{25}{3}$

i.e. $\;$ $a = r + \dfrac{25}{3}$ $\;\;\; \cdots \; (2)$

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

$\left(r + \dfrac{25}{3}\right) \times \left(1 + r\right) = 15$

i.e. $\;$ $r^2 + \dfrac{28 r}{3} + \dfrac{25}{3} = 15$

i.e. $\;$ $3r^2 + 28 r - 20 = 0$

i.e. $\;$ $3 r^2 + 30 r - 2 r - 20 = 0$

i.e. $\;$ $3r \left(r + 10\right) - 2 \left(r + 10\right) = 0$

i.e. $\;$ $\left(r + 10\right) \left(3r - 2\right) = 0$

i.e. $\;$ $r = -10$ $\;$ or $\;$ $r = \dfrac{2}{3}$

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

when $\;$ $r = -10$, $\;$ $a = -10 + \dfrac{25}{3} = \dfrac{-5}{3}$

and when $\;$ $r = \dfrac{2}{3}$, $\;$ $a = \dfrac{2}{3} + \dfrac{25}{3} = 9$

Now, fourth term of the G.P $= t_4 = a \cdot r^3$

$\therefore \;$ when $\;$ $a = \dfrac{-5}{3}$, $\;$ $r = -10$, $\;$ $t_4 = \dfrac{-5}{3} \times \left(-10\right)^3 = \dfrac{5000}{3}$

and when $\;$ $a = 9$, $\;$ $r = \dfrac{2}{3}$, $\;$ $t_4 = 9 \times \left(\dfrac{2}{3}\right)^3 = \dfrac{8}{3}$

Algebra - Geometric Progressions

The sum of the first four terms of a geometric progression (G.P) is $30$ and the sum of the next four terms is $480$. Find the sum of the first twelve terms.

Let the first term of the G.P be $= t_1 = a$

and the common ratio be $= r$

Sum of $n$ terms of G.P $= S_n = \dfrac{a \left(r^n - 1\right)}{r - 1}$

Given: $\;$ Sum of first four terms of G.P $= S_4 = 30$

i.e. $\;$ $S_4 = \dfrac{a \left(r^4 - 1\right)}{r - 1} = 30$ $\;\;\; \cdots \; (1)$

And: $\;$ Sum of next four terms $= 480$

i.e. $\;$ Sum of first $8$ terms $-$ Sum of first $4$ terms $= 480$

i.e. $\;$ $S_8 - S_4 = 480$

i.e. $\;$ $S_8 = 480 + S_4 = 480 + 30 = 510$ $\;\;\; \left[\because \; S_4 = 30\right]$

i.e. $\;$ $S_8 = \dfrac{a \left(r^8 - 1\right)}{r - 1} = 510$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$,

$\dfrac{S_8}{S_4} = \dfrac{r^8 - 1}{r^4 - 1} = \dfrac{510}{30} = 17$

i.e. $\;$ $\dfrac{\left(r^4\right)^2 - 1^2}{r^4 - 1} = 17$

i.e. $\;$ $\dfrac{\left(r^4 + 1\right) \left(r^4 - 1\right)}{r^4 - 1} = 17$

i.e. $\;$ $r^4 + 1 = 17$

i.e. $\;$ $r^4 = 16 = 2^4$ $\implies$ $r = 2$

Substituting the value of $r$ in equation $(1)$ gives

$30 = \dfrac{a \left(2^4 - 1\right)}{2 - 1}$

i.e. $\;$ $30 = 15 \times a$ $\implies$ $a = 2$

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

$= S_{12} = \dfrac{2 \times \left(2^{12} - 1\right)}{2 - 1} = 2 \times 4095 = 8190$

Algebra - Geometric Progressions

The difference between the fourth and the first term of a geometric progression (G.P) is $52$ and the sum of the first three terms is $26$. Calculate the sum of the first six terms of the progression.

Let the first term of the G.P be $= t_1 = a$

and the common ratio be $= r$

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

$\therefore \;$ $4^{th}$ term of G.P $= t_4 = a \cdot r^3$

Sum of $n$ terms of G.P $= S_n = \dfrac{a \left(r^n - 1\right)}{r - 1}$

Given: $\;$ $t_4 - t_1 = 52$

i.e. $\;$ $a \cdot r^3 - a = 52$

i.e. $\;$ $a \left(r^3 - 1\right) = 52$ $\;\;\; \cdots \; (1)$

And: $\;$ $S_3 = 26$

i.e. $\;$ $\dfrac{a \left(r^3 - 1\right)}{r - 1} = 26$ $\;\;\; \cdots \; (2)$

In view of equation $(1)$, equation $(2)$ becomes

$\dfrac{52}{r - 1} = 26$

i.e. $\;$ $r - 1 = 2$ $\implies$ $r = 3$

Substituting the value of $r$ in equation $(1)$ gives

$a \left(3^3 - 1\right) = 52$ $\implies$ $a = \dfrac{52}{26} = 2$

$\therefore \;$ Sum of the first six terms of G.P

$= S_6 = \dfrac{a \left(r^6 - 1\right)}{r - 1} = \dfrac{2 \times \left(3^6 - 1\right)}{3 - 1} = 728$

Algebra - Geometric Progressions

The fourth term of a geometric progression (G.P) exceeds the second term by $24$ and the sum of the second and the third terms is $6$. Find the progression.

Let the first term of the G.P be $= t_1 = a$

and the common ratio be $= r$

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

$\therefore \;$ $2^{nd}$ term of G.P $= t_2 = a \cdot r$

$3^{rd}$ term of G.P $= t_3 = a \cdot r^2$

$4^{th}$ term of G.P $= t_4 = a \cdot r^3$

Given: $\;$ $t_4 = t_2 + 24$

i.e. $\;$ $a \cdot r^3 = a \cdot r + 24$

i.e. $\;$ $a \cdot r \cdot \left(r^2 - 1\right) = 24$

i.e. $\;$ $a \cdot r \cdot \left(r + 1\right) \left(r - 1\right) = 24$

i.e. $\;$ $a \cdot r \cdot \left(r + 1\right) = \dfrac{24}{r - 1}$ $\; \; \; \cdots \; (1)$

And: $\;$ $t_2 + t_3 = 6$

i.e. $\;$ $a \cdot r + a \cdot r^2 = 6$

i.e. $\;$ $a \cdot r \cdot \left(r + 1\right) = 6$

i.e. $\;$ $\dfrac{24}{r - 1} = 6$ $\;\;\;$ [in view of equation $(1)$]

i.e. $\;$ $r - 1 = 4$ $\implies$ $r = 5$

Substituting the value of $r$ in equation $(1)$ gives

$5 \times a \times \left(5 + 1\right) = \dfrac{24}{5 - 1}$

i.e. $\;$ $5 \times 6 \times a = 6$

$\implies$ $a = \dfrac{1}{5}$

$\therefore \;$ The G.P is $\;\;$ $a, \; a \cdot r, \; a \cdot r^2, \; a \cdot r^3, \cdots$

i.e. $\;$ $\dfrac{1}{5}, \; \dfrac{1}{5} \times 5, \; \dfrac{1}{5} \times 5^2, \; \dfrac{1}{5} \times 5^3, \cdots$

i.e. $\;$ $\dfrac{1}{5}, \; 1, \; 5, \; 25, \cdots$

Algebra - Arithmetic Progressions

The fourth term of an arithmetic progression (A.P) is $4$. At what value of the difference of the progression is the sum of the pairwise products of the first three terms of the progression the least?

Let the first term of the A.P be $= t_1 = a$

and the common difference be $= d$

$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right) d$

$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$

$3^{rd}$ term of A.P $= t_3 = a + 2d$

$4^{th}$ term of A.P $= t_4 = a + 3d$

Given: $\;$ $t_4 = a + 3d = 4$

i.e. $\;$ $a = 4 - 3d$ $\;\;\; \cdots \; (1)$

And: $\;$ Sum of pairwise products of the first three terms of the progression is the least

Let $\;$ $S =$ sum of pairwise products of the first three terms of A.P

i.e. $\;$ $S = t_1 \times t_2 + t_2 \times t_3 + t_3 \times t_1$

i.e. $\;$ $S = a \times \left(a + d\right) + \left(a + d\right) \left(a + 2d\right) + \left(a + 2d\right) \times a$

i.e. $\;$ $S = a^2 + ad + a^2 + 3ad + 2d^2 + a^2 + 2ad$

i.e. $\;$ $S = 3a^2 + 6ad + 2d^2$ $\;\;\; \cdots \; (2)$

In view of equation $(1)$, equation $(2)$ becomes

$S = 3 \times \left(4 - 3d\right)^2 + 6 \times \left(4 - 3d\right) \times d + 2 d^2$

i.e. $\;$ $S = 3 \times \left(16 + 9 d^2 - 24 d\right) + 24 d - 18 d^2 + 2 d^2$

i.e. $\;$ $S = 48 + 27 d^2 - 72 d + 24 d - 16 d^2$

i.e. $\;$ $48 - 48 d + 11 d^2$ $\;\;\; \cdots \; (3)$

For $S$ to be least, $\;$ $\dfrac{d \left(S\right)}{d \left(d\right)} = 0$

$\therefore \;$ We have from equation $(3)$,

$\dfrac{d \left(S\right)}{d \left(d\right)} = 0 - 48 + 22d = 0$

i.e. $\;$ $d = \dfrac{48}{22} = \dfrac{24}{11}$

$\therefore \;$ $S$ will be least when $d = \dfrac{24}{11}$