Algebra - Arithmetic Progressions

Prove that the numbers $\dfrac{1}{\log_3 2}$, $\dfrac{1}{\log_6 2}$ and $\dfrac{1}{\log_{12} 2}$ form an arithmetic progression (A.P).

To prove that (TPT) $\;$ $\dfrac{1}{\log_3 2}$, $\;$ $\dfrac{1}{\log_6 2}$ $\;$ and $\;$ $\dfrac{1}{\log_{12} 2}$ $\;$ form an A.P.

i.e. $\;$ TPT $\;$ $\log_2 3$, $\;$ $\log_2 6$ $\;$ and $\;$ $\log_2 12$ $\;$ form an A.P.

i.e. $\;$ TPT $\;$ $\log_2 6 - \log_2 3 = \log_2 12 - \log_2 6$

[Three numbers $\; a, \; b, \; c \;$ are in A.P. if $\;$ $b - a = c - b$]

i.e. $\;$ TPT $\;$ $\log_2 \left(\dfrac{6}{3}\right) = \log_2 \left(\dfrac{12}{6}\right)$

i.e. $\;$ TPT $\;$ $\log_2 2 = \log_2 2$ $\;\;$ which is true.

i.e. $\;$ TPT $\;$ $1 = 1$ $\;$ which is true.

$\implies$ The three numbers are in A.P.

Algebra - Arithmetic Progressions

Three numbers form an arithmetic progression (A.P). The sum of the numbers is equal to $3$ and the sum of their cubes is equal to $4$. Find the numbers.

Let the first term of the A.P be $= a$ $\;$ and the common difference $= d$

Let the three terms of A.P be $\;$ $a - d, \; a, \; a + d$

Given: $\;$ Sum of the three numbers is equal to $3$

i.e. $\;$ $a -d + a + a + d = 3$

i.e. $\;$ $3a = 3$ $\implies$ $a = 1$

And: $\;$ Sum of the cubes of these three numbers is equal to $4$

i.e. $\;$ $\left(a - d\right)^3 + a^3 + \left(a + d\right)^3 = 4$

i.e. $\;$ $\left(1 - d\right)^3 + 1^3 + \left(1 + d\right)^3 = 4$ $\;\;\;$ $\left[\because \; a = 1\right]$

i.e. $\;$ $\left(1 - d + 1 + d\right) \left[\left(1 - d\right)^2 - \left(1 - d\right) \left(1 + d\right) + \left(1 + d\right)^2\right] = 3$

$\left[\because \; p^3 + q^3 = \left(p + q\right) \left(p^2 - pq + q^2\right)\right]$

i.e. $\;$ $2 \left[1 - 2d + d^2 - 1 + d^2 + 1 + 2d + d^2\right] = 3$

i.e. $\;$ $3 d^2 + 1 = \dfrac{3}{2}$

i.e. $\;$ $3 d^2 = \dfrac{1}{2}$

i.e. $\;$ $d^2 = \dfrac{1}{6}$ $\implies$ $d = \pm \dfrac{1}{\sqrt{6}}$

When $\;$ $a = 1$ $\;$ and $\;$ $d = \dfrac{+ 1}{\sqrt{6}}$, $\;$ the three numbers are $\;$ $1 - \dfrac{1}{\sqrt{6}}$, $\;$ $1$, $\;$ $1 + \dfrac{1}{\sqrt{6}}$

When $\;$ $a = 1$ $\;$ and $\;$ $d = \dfrac{- 1}{\sqrt{6}}$, $\;$ the three numbers are $\;$ $1 + \dfrac{1}{\sqrt{6}}$, $\;$ $1$, $\;$ $1 - \dfrac{1}{\sqrt{6}}$

$\therefore \;$ The three numbers in A.P are $\;$ $1 + \dfrac{1}{\sqrt{6}}, \; 1, \; 1 - \dfrac{1}{\sqrt{6}}$

Algebra - Arithmetic Progressions

Each of the two triplets of numbers $\log a$, $\log b$, $\log c$ $\;$ and $\;$ $\log a - \log 2b$, $\log 2b - \log 3c$, $\log 3c - \log a$ is an arithmetic progression (A.P). Can the numbers $a$, $b$ and $c$ be the lengths of the sides of a triangle? If they can, then what triangle is it? Find the angles of the triangle provided that it exists.

Given: $\;$ $\log a$, $\log b$, $\log c$ are in A.P

$\implies$ $2 \log b = \log a + \log c$

i.e. $\log b^2 = \log ac$

i.e. $b^2 = ac$ $\;\;\; \cdots \; (1)$

And: $\;$ $\log a - \log 2b$, $\log 2b - \log 3c$, $\log 3c - \log a$ are in A.P

$\implies$ $2 \left(\log 2b - \log 3c\right) = \log a - \log 2b + \log 3c - \log a$

i.e. $\;$ $2 \times \log \left(\dfrac{2b}{3c}\right) = \log \left(\dfrac{3c}{2b}\right)$

i.e. $\;$ $\log \left(\dfrac{2b}{3c}\right)^2 = \log \left(\dfrac{3c}{2b}\right)$

i.e. $\;$ $\left(\dfrac{2b}{3c}\right)^2 = \dfrac{3c}{2b}$

i.e. $\;$ $\left(2b\right)^3 = \left(3c\right)^3$

i.e. $\;$ $2b = 3c$ $\implies$ $b = \dfrac{3c}{2} = 1.5 c$ $\;\;\; \cdots \; (2)$

$\therefore \;$ In view of equation $(2)$, we have from equation $(1)$

$\dfrac{9c^2}{4} = ac$ $\implies$ $a = \dfrac{9c}{4} = 2.25c$ $\;\;\; \cdots \; (3)$

$\therefore \;$ The three numbers that satisfy the given conditions are

$a = \dfrac{9c}{4} = 2.25c$, $\;$ $b = \dfrac{3c}{2} = 1.5c$, $\;$ $c \; \left(c > 0\right)$

Check the existance of a triangle with sides $a$, $b$ and $c$:

For $a$, $b$ and $c$ to form the sides of a triangle, the condition to be satisfied is

$a + b >c$ $\;\;\; \cdots \; (4a)$, $\;$ $b + c > a$ $\;\;\; \cdots \; (4b)$, $\;$ $c + a > b$ $\;\;\; \cdots \; (4c)$

In view of equations $(2)$ and $(3)$,

equation $(4a)$ $\implies$ $2.25 c + 1.5c = 3.75 c > c$ $\;$ which is true

equation $(4b)$ $\implies$ $1.5 c + c = 2.5 c > 2.25c$ $\;$ which is true

and equation $(4c)$ $\implies$ $c + 2.25 c = 3.25 c > 1.5c$ $\;$ which is true

$\implies$ A triangle can be formed with sides $a$, $b$ and $c$.

Type of triangle:

From equation $(2)$, $\;$ $b^2 = \dfrac{9c^2}{4} = 2.25 c^2$ $\;\;\; \cdots \; (5a)$

From equation $(3)$, $\;$ $a^2 = \dfrac{81 c^2}{16} = 5.0625 c^2$ $\;\;\; \cdots \; (5b)$

$\therefore \;$ We have from equations $(5a)$ and $(5b)$

$a^2 = 5.0625 c^2 > b^2 + c^2 = 2.25 c^2 + c^2 = 3.25 c^2$

$\implies$ The triangle with sides $a$, $b$ and $c$ is an obtuse triangle.

Angles of the triangle:

By cosine rule,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{\dfrac{9c^2}{4} + c^2 - \dfrac{81 c^2}{16}}{2 \times \dfrac{3c}{2} \times c} = \dfrac{-29}{48}$

$\implies$ $A = \cos^{-1} \left(\dfrac{-29}{48}\right) = \pi - \cos^{-1} \left(\dfrac{29}{48}\right)$

$\cos B = \dfrac{c^2 + a^2 - b^2}{2ca} = \dfrac{c^2 + \dfrac{81 c^2}{16} - \dfrac{9c^2}{4}}{2 \times c \times \dfrac{9c}{4}} = \dfrac{61}{72}$

$\implies$ $B = \cos^{-1} \left(\dfrac{61}{72}\right)$

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{\dfrac{81 c^2}{16} + \dfrac{9c^2}{4} - c^2}{2 \times \dfrac{9c}{4} \times \dfrac{3c}{2}} = \dfrac{101}{108}$

$\implies$ $C = \cos^{-1} \left(\dfrac{101}{108}\right)$

$\therefore \;$ The angles of the triangle are

$\angle A = \pi - \cos^{-1} \left(\dfrac{29}{48}\right)$, $\;$ $\angle B = \cos^{-1} \left(\dfrac{61}{72}\right)$, $\;$ $\angle C = \cos^{-1} \left(\dfrac{101}{108}\right)$

Algebra - Arithmetic Progressions

Two arithmetic progressions (A.Ps) contain the same number of terms. The ratio of the last term of the first progression to the first term of the second is equal to the ratio of the last term of the second progression to the first term of the first progression and is equal to $4$. The ratio of the sum of the first progression to that of the second is $2$. Find the ratio of the differences of the progressions.

Let the number of terms in the two A.Ps be $= n$

Let the $n^{th}$ term be the last term of both the progressions.

For first A.P:

$1^{st}$ term $= t_{11} = a_{11}$, $\;$ common difference $= d_1$

Sum of all terms $= S_{1n} = \dfrac{n}{2} \left[2a_{11} + \left(n - 1\right) d_1\right]$

For second A.P:

$1^{st}$ term $= t_{21} = a_{21}$, $\;$ common difference $= d_2$

Sum of all terms $= S_{2n} = \dfrac{n}{2} \left[2a_{21} + \left(n - 1\right) d_2\right]$

To find: $\;$ $\dfrac{d_1}{d_2}$

Given: $\;$ $\dfrac{t_{1n}}{t_{21}} = \dfrac{t_{2n}}{t_{11}} = 4$

i.e. $\;$ $\dfrac{a_{11} + \left(n - 1\right) d_1}{a_{21}} = \dfrac{a_{21} + \left(n - 1\right) d_2}{a_{11}} = 4$ $\;\;\; \cdots \; (1)$

$\implies$ $a_{11} + \left(n - 1\right) d_1 = 4 a_{21}$ $\;\;\; \cdots \; (2)$

and $\;$ $a_{21} + \left(n - 1\right)d_2 = 4 a_{11}$ $\;\;\; \cdots \; (3)$

Also: $\;$ $\dfrac{S_{1n}}{S_{2n}} = 2$

i.e. $\;$ $\dfrac{\dfrac{n}{2} \left[2a_{11} + \left(n - 1\right) d_1\right]}{\dfrac{n}{2} \left[2a_{21} + \left(n - 1\right) d_2\right]} = 2$

i.e. $\;$ $\dfrac{2a_{11} + \left(n - 1\right) d_1}{2 a_{21} + \left(n - 1\right) d_2} = 2$

i.e. $\;$ $\dfrac{a_{11} + a_{11} + \left(n - 1\right) d_1}{a_{21} + a_{21} + \left(n -1\right)d_2} = 2$

i.e. $\;$ $\dfrac{a_{11} + 4a_{21}}{a_{21} + 4 a_{11}} = 2$ $\;\;$ [By equations $(2)$ and $(3)$]

i.e. $\;$ $a_{11} + 4 a_{21} = 2 a_{21} + 8 a_{11}$

i.e. $\;$ $2 a_{21} = 7 a_{11}$

i.e. $\;$ $a_{21} = \dfrac{7}{2} a_{11}$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(3)$ becomes

$\dfrac{7}{2} a_{11} + \left(n - 1\right) d_2 = 4 a_{11}$

i.e. $\;$ $\left(n - 1\right) d_2 = 4a_{11} - \dfrac{7}{2} a_{11} = \dfrac{a_{11}}{2}$ $\;\;\; \cdots \; (5)$

In view of equation $(4)$, equation $(2)$ becomes

$a_{11} + \left(n - 1\right) d_1 = 4 \times \dfrac{7}{2} a_{11} = 14 a_{11}$

i.e. $\;$ $\left(n - 1\right) d_1 = 13 a_{11}$ $\;\;\; \cdots \; (6)$

$\therefore \;$ We have from equations $(5)$ and $(6)$

$\dfrac{\left(n - 1\right) d_1}{\left(n - 1\right) d_2} = \dfrac{13 a_{11}}{a_{11} / 2}$

$\implies$ $\dfrac{d_1}{d_2} = 26$

Algebra - Arithmetic Progressions

In an arithmetic progression (A.P), $a_7 = 9$. At what value of its difference is the product $a_1 \cdot a_2 \cdot a_7$ the least?

Let the first term of A.P $= a_1$ and the common difference $= d$

$n^{th}$ term of A.P $= a_n = a_1 + \left(n - 1\right) d$

$\therefore \;$ $2^{nd}$ term of A.P $= a_2 = a_1 + d$ $\;\;\; \cdots \; (1)$

and $\;$ $7^{th}$ term of A.P $= a_7 = a_1 + 6d$

Given: $\;$ $a_7 = 9$ $\;\;\; \cdots \; (2)$

i.e. $\;$ $a_1 + 6d = 9$ $\implies$ $a_1 = 9 - 6d$ $\;\;\; \cdots \; (3)$

In view of equation $(3)$, equation $(1)$ becomes

$a_2 = 9 - 6d + d = 9 - 5d$ $\;\;\; \cdots \; (4)$

From equations $(2)$, $(3)$ and $(4)$, the product

$\begin{aligned} P = a_1 \times a_2 \times a_7 & = \left(9 - 6d\right) \times \left(9 - 5d\right) \times 9 \\\\ & = 9 \times \left(81 - 99d + 30 d^2\right) \end{aligned}$

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

i.e. $\;$ $\dfrac{d\left(P\right)}{d \left(d\right)} = 9 \times \left(-99 + 60d\right) = 0$

$\implies$ $d = \dfrac{99}{60} = \dfrac{33}{20}$

$\therefore \;$ When the common difference of the A.P is $d = \dfrac{33}{20}$, the product $a_1 \times a_2 \times a_7$ is the least.

Algebra - Arithmetic Progressions

The product of the third by the sixth term of an arithmetic progression (A.P) is $406$. The division of the ninth term of the progression by the fourth term gives a quotient $2$ and a remainder $6$. Find the first term and the difference of the progression.

Let the first term of A.P $= t_1 = a$ and the common difference $= d$

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

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

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

$6^{th}$ term of A.P $= t_6 = a + 5d$;

$9^{th}$ term of A.P $= t_9 = a + 8d$;

Given: $\;$ $t_3 \times t_6 = 406$

i.e. $\;$ $\left(a + 2d\right) \left(a + 5d\right) = 406$

i.e. $\;$ $a^2 + 7ad + 10d^2 = 406$ $\;\;\; \cdots \; (1)$

Also: $\;$ $t_9 \div t_4$ gives a quotient $2$ and a remainder $6$

i.e. $\;$ $\dfrac{t_9}{t_4} = 2 + \dfrac{6}{t_4}$

i.e. $\;$ $t_9 = 2 t_4 + 6$

i.e. $\;$ $a + 8d = 2 \times \left(a + 3d\right) + 6$

i.e. $\;$ $a + 8d = 2a + 6d + 6$

i.e. $\;$ $a = 2d - 6$ $\;\;\; \cdots \; (2)$

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

$\left(2d - 6\right)^2 + 7 \times \left(2d - 6\right) \times d + 10d^2 = 406$

i.e. $\;$ $4d^2 - 24d + 36 + 14d^2 - 42d + 10d^2 = 406$

i.e. $\;$ $28 d^2 - 66d - 370 = 0$

i.e. $\;$ $14 d^2 - 33 d - 185 = 0$

i.e. $\;$ $d = \dfrac{33 \pm \sqrt{33^2 - 4 \times 14 \times \left(-185\right)}}{2 \times 14}$

i.e. $\;$ $d = \dfrac{33 \pm \sqrt{1089 + 10360}}{28} = \dfrac{33 \pm \sqrt{11449}}{28}$

i.e. $\;$ $d = \dfrac{33}{28} \pm \dfrac{107}{28}$

i.e. $\;$ $d = 5$ $\;$ or $\;$ $d = \dfrac{-37}{14}$

When $\;$ $d = 5$, $\;$ we have from equation $(2)$, $\;$ $a = \left(2 \times 5\right) - 6 = 4$

When $\;$ $d = \dfrac{-37}{14}$, $\;$ we have from equation $(2)$, $\;$ $a = \left[2 \times \left(\dfrac{-37}{14}\right)\right] - 6 = \dfrac{-79}{7}$

Algebra - Arithmetic Progressions

An arithmetic progression (A.P) consists of $12$ terms whose sum is $354$. The ratio of the sum of the even terms to the sum of the odd terms is $32 : 27$. Find the common difference of the progression.

Let the first term of A.P $= t_1 = a$ and the common difference $= d$

Sum to $n$ terms of an A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$

Given: Number of terms of A.P $= n = 12$; $\;\;$ $S_n = 354$

i.e. $\;$ $\dfrac{12}{2} \times \left[2a + 11d\right] = 354$

i.e. $\;$ $2a + 11d = 59$ $\;\;\; \cdots \; (1)$

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

Even terms of A.P $= t_2, \; t_4, \; t_6, \cdots, t_{12}$ $\;$ i.e. $\;$ $6$ terms.

For even terms, first term $= t_2 = a + d$ $\;$ and last term $= t_{12} = a + 11d$

$\therefore \;$ Sum of even terms of A.P $= S_{\text{even}} = \dfrac{t_2 + t_{12}}{2}$

i.e. $\;$ $S_{\text{even}} = \dfrac{a + d + a + 11d}{2} = a + 6d$

Odd terms of A.P $= t_1, \; t_3, \; t_5, \cdots, t_{11}$ $\;$ i.e. $\;$ $6$ terms.

For odd terms, first term $= t_1 = a$ $\;$ and last term $= t_{11} = a + 10d$

$\therefore \;$ Sum of odd terms of A.P $= S_{\text{odd}} = \dfrac{t_1 + t_{11}}{2}$

i.e. $\;$ $S_{\text{odd}} = \dfrac{a + a + 10d}{2} = a + 5d$

Given: $\dfrac{S_{\text{even}}}{S_{\text{odd}}} = \dfrac{32}{27}$

i.e. $\;$ $\dfrac{a + 6d}{a + 5d} = \dfrac{32}{27}$

i.e. $\;$ $27a + 162 d = 32a + 160d$

i.e. $\;$ $2d = 5a$ $\implies$ $a = \dfrac{2d}{5}$

Substituting $a = \dfrac{2d}{5}$ in equation $(1)$ gives

$2 \times \dfrac{2d}{5} + 11d = 59$

i.e. $\;$ $\dfrac{59 d}{5} = 59$ $\implies$ $d = 5$

i.e. $\;$ The common difference of the given A.P is $d = 5$