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 $d = 5$

Algebra - Arithmetic Progressions

Solve the equation: $\;\;\;$ $5^2 \times 5^4 \times 5^6 \times \cdots \times 5^{2x} = \left(0.04\right)^{-28}$

Given equation: $\;\;\;$ $5^2 \times 5^4 \times 5^6 \times \cdots \times 5^{2x} = \left(0.04\right)^{-28}$

i.e. $\;$ $5^{\left[2 + 4 + 6 + \cdots + 2x\right]} = \left(\dfrac{1}{25}\right)^{-28}$

i.e. $\;$ $5^{\left[2 + 4 + 6 + \cdots + 2x\right]} = \left(5^{-2}\right)^{-28} = 5^{56}$

$\implies$ $2 + 4 + 6 + \cdots + 2x = 56$ $\;\;\; \cdots \; (1)$

Now, $\;$ $2 + 4 + 6 + \cdots + 2x$ $\;$ is an arithmetic progression (A.P) with

first term $= a = 2$,

common difference $= d = 2$,

$n^{th}$ term $= t_n = 2x$ $\;$ and

sum to $n$ terms $= S_n = 56$

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

i.e. $\;$ $2x = 2 + \left(n - 1\right) \times 2$

i.e. $\;$ $x = 1 + n - 1$ $\implies$ $x = n$ $\;\;\; \cdots \; (2)$

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

i.e. $\;$ $56 = \dfrac{n}{2} \times \left[2 \times 2 + \left(n - 1\right) \times 2\right]$

i.e. $\;$ $112 = n \left[4 + 2n - 2\right]$

i.e. $\;$ $2n^2 + 2n - 112 = 0$

i.e. $\;$ $n^2 + n - 56 = 0$

i.e. $\;$ $\left(n - 7\right) \left(n + 8\right) = 0$

i.e. $\;$ $n - 7 = 0$ $\;$ or $\;$ $n + 8 = 0$

i.e. $\;$ $n = 7$ $\;$ or $\;$ $n = -8$

Since the number of terms of an A.P cannot be negative,

$\implies$ $n = 7$

$\therefore \;$ By equation $(2)$, $\;$ $x = 7$

Algebra - Arithmetic Progressions

Put five terms between the numbers $1$ and $1.3$ so that together with the given terms they will form an arithmetic progression (A.P).

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

Let the common difference of A.P $= d$

Five terms are to be put between the numbers $1$ and $1.3$

$\implies$ The $1^{st}$ term of the required A.P $= a = 1$

and the $7^{th}$ term of the required A.P $= t_7 = 1.3$

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

$\therefore \;$ $7^{th}$ term of A.P $= t_7 = a + 6d = 1.3$ $\;\;\; \cdots \; (1)$

Substituting $a = 1$ in equation $(1)$ gives $\;\;$ $1 + 6d = 1.3$

i.e. $\;$ $6d = 0.3$ $\implies$ $d = 0.05$

$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d = 1 + 0.05 = 1.05$

$3^{rd}$ term of A.P $= t_3 = t_2 + d = 1.05 + 0.05 = 1.10$

$4^{th}$ term of A.P $= t_4 = t_3 + d = 1.10 + 0.05 = 1.15$

$5^{th}$ term of A.P $= t_5 = t_4 + d = 1.15 + 0.05 = 1.20$

$6^{th}$ term of A.P $= t_6 = t_5 + d = 1.20 + 0.05 = 1.25$

$\therefore \;$ The five terms to be put between the numbers $1$ and $1.3$ so that together with the given terms they will form an A.P are: $\;\;$ $1.05, \; 1.10, \; 1.15, \; 1.20 \;$ and $\; 1.25$

Algebra - Arithmetic Progressions

How may terms of an arithmetic progression (A.P) must be taken for their sum to equal $91$, if its third term is $9$ and the difference between the seventh and the second term is $20$?

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

Let the common difference of A.P $= d$

Let the numebr of terms required for the sum of terms of the A.P to equal $91$ be $= n$

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

$n^{th}$ 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 $= t_3 = a + 2d$

$7^{th}$ term of A.P $= t_7 = a + 6d$

Given: $\;$ $S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right] = 91$

i.e. $\;$ $n \left[2a + \left(n - 1\right) d\right] = 182$ $\;\;\; \cdots \; (1)$

And $\;$ $t_3 = a + 2d = 9$ $\;\;\; \cdots \; (2)$

And $\;$ $t_7 - t_2 = 20$

i.e. $\;$ $a + 6d - \left(a +d\right) = 20$

i.e. $\;$ $5 d = 20$ $\implies$ $d = 4$

Substituting $d = 4$ in equation $(2)$ gives

$a + 8 = 9$ $\implies$ $a = 1$

Substituting $a = 1, \; d = 4$ in equation $(1)$ gives

$n \left[2 \times 1 + \left(n - 1\right) \times 4\right] = 182$

i.e. $\;$ $n \left(2 + 4n - 4\right) = 182$

i.e. $\;$ $4 n^2 - 2n = 182$

i.e. $\;$ $2n^2 - n - 91 = 0$

i.e. $\;$ $2 n^2 - 14 n + 13 n - 91 = 0$

i.e. $\;$ $2n \left(n - 7\right) + 13 \left(n - 7\right) = 0$

i.e. $\;$ $\left(2n + 13\right) \left(n - 7 = 0\right)$

i.e. $\;$ $n = \dfrac{-13}{2}$ $\;$ or $\;$ $n = 7$

Since the number of terms of an A.P cannot be negative,

$\therefore \;$ the required number of terms of A.P $= n = 7$

Algebra - Arithmetic Progressions

The sum of the second and the fifth term of an arithmetic progression (A.P) is $8$ and that of the third and the seventh term is $14$. Find the progression.

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

Let the common difference of A.P $= d$

$n^{th}$ 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 $= t_3 = a + 2d$

$5^{th}$ term of A.P $= t_5 = a + 4d$

$7^{th}$ term of A.P $= t_7 = a + 6d$

Given: $\;$ $t_2 + t_5 = 8$

i.e. $\;$ $a + d + a + 4d = 8$

i.e. $\;$ $2a + 5d = 8$ $\;\;\; \cdots \; (1)$

And $\;$ $t_3 + t_7 = 14$

i.e. $\;$ $a + 2d = a + 6d = 14$

i.e. $\;$ $2a + 8d = 14$ $\;\;\; \cdots \; (2)$

Subtracting equations $(1)$ and $(2)$ gives

$3d = 6$ $\implies$ $d = 2$

Substituting $d = 2$ in equation $(1)$ gives

$2a + 5 \times 2 = 8$

i.e. $\;$ $2a = -2$ $\implies$ $a = -1$

$\therefore \;$ The required A.P is $\;$ $a, \; a+d, \; a+2d, \; a+3d, \; a+4d, \; a+5d, \; a+6d, \; \cdots$

i.e. $\;$ $-1, \; 1, \; 3, \; 5, \; 7, \; 9, \; 11, \; \cdots$

Algebra - Arithmetic Progressions

The sum of the squares of the fifth and the eleventh term of an arithmetic progression (A.P) is $3$ and the product of the second by the fourteenth term is equal to $k$. Find the product of the first by the fifteenth term of the progression.

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

Let the common difference of A.P $= d$

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

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

$5^{th}$ term of A.P $= t_5 = a + 4d$

$11^{th}$ term of A.P $= t_{11} = a + 10d$

$14^{th}$ term of A.P $= t_{14} = a + 13d$

Given: $\;$ $\left(t_5\right)^2 + \left(t_{11}\right)^2 = 3$

i.e. $\;$ $\left(a + 4d\right)^2 + \left(a + 10d\right)^2 = 3$

i.e. $\;$ $a^2 + 16d^2 + 8ad + a^2 + 100 d^2 + 20 ad = 3 $

i.e. $\;$ $2a^2 + 116 d^2 + 28 ad = 3$ $\;\;\; \cdots \; (1)$

And $\;$ $t_2 \times t_{14} = k$

i.e. $\;$ $\left(a + d\right) \left(a + 13d\right) = k$

i.e. $\;$ $a^2 + 13ad + ad + 13 d^2 = k$

i.e. $\;$ $a^2 + 13 d^2 + 14ad = k$ $\;\;\; \cdots \; (2)$

Multiplying equation $(2)$ with $2$ gives

$2a^2 + 26d^2 + 28ad = 2k$ $\;\;\; \cdots \; (3)$

Subtracting equations $(1)$ and $(3)$ gives

$90 d^2 = 3 - 2k$

i.e. $\;$ $d^2 = \dfrac{3 - 2k}{90}$ $\;\;\; \cdots \; (4)$

To find: $\;$ $t_1 \times t_{15} = a \times \left(a + 14 d\right) = a^2 + 14ad$ $\;\;\; \cdots \; (5)$

We have from equation $(2)$, $\;$ $a^2 + 14ad = k - 13 d^2$

$\begin{aligned} \implies t_1 \times t_{15} & = k - 13 d^2 \\\\ & = k - 13 \times \left(\dfrac{3 - 2k}{90}\right) \;\;\; \left[\text{by equation (4)}\right] \\\\ & = \dfrac{90 k - 39 + 26 k}{90} \\\\ & = \dfrac{116 k - 39}{90} \end{aligned}$

Algebra - Arithmetic Progressions

The sum of the first and the fifth term of an arithmetic progression (A.P) is $26$ and the product of the second by the fourth term is $160$. Find the sum of the first six terms of the progression.

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

Let the common difference of A.P $= d$

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

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

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

$5^{th}$ term of A.P $= t_5 = a + 4d$

Given: $\;$ $t_1 + t_5 = 26$

i.e. $\;$ $a + a + 4d = 26$

i.e. $\;$ $2a + 4d = 26$

i.e. $\;$ $a + 2d = 13$

i.e. $\;$ $a = 13 - 2d$ $\;\;\; \cdots \; (1)$

And $\;$ $t_2 \times t_4 = 160$

i.e. $\;$ $\left(a + d\right) \left(a + 3d\right) = 160$ $\;\;\; \cdots \; (2)$

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

$\left(13 - 2d + d\right) \left(13 - 2d + 3d\right) = 160$

i.e. $\;$ $\left(13 - d\right) \left(13 + d\right) = 160$

i.e. $\;$ $169 - d^2 = 160$

i.e. $\;$ $d^2 = 9$ $\implies$ $d = \pm 3$

When $\;$ $d = +3$, $\;$ we have from equation $(1)$, $\;$ $a = 13 - 6 = 7$

When $\;$ $d = -3$, $\;$ we have from equation $(1)$, $\;$ $a = 13 + 6 = 19$

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

$\therefore \;$ Sum of first six terms of A.P $= S_6 = \dfrac{6 \times \left[2a + 5d\right]}{2} = 3 \left[2a + 5d\right]$

Case 1: $\;$ When $a = 7, \; d = +3$, $\;$ $S_6 = 3 \times \left[2 \times 7 + 5 \times 3\right] = 87$

Case 2: $\;$ When $a = 19, \; d = -3$, $\;$ $S_6 = 3 \times \left[2 \times 19 - 5 \times 3\right] = 69$

Algebra - Arithmetic Progressions

Find the first term and the common difference of an arithmetic progression (A.P) if the sum of its first five even terms is equal to $15$ and the sum of the first three terms is equal to $-3$.

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

Let the common difference of A.P $= d$

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

Given: Sum of the first five even terms of A.P $= 15$

i.e. $\;$ $t_2 + t_4 + t_6 + t_8 + t_{10} = 15$

i.e. $\;$ $\left(a + d\right) + \left(a + 3d\right) + \left(a + 5d\right) + \left(a + 7d\right) + \left(a + 9d\right) = 15$

i.e. $\;$ $5a + 25d = 15$

i.e. $\;$ $a + 5d = 3$ $\;\;\; \cdots \; (1)$

And, sum of the first three terms $= -3$

i.e. $\;$ $t_1 + t_2 + t_3 = -3$

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

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

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

Subtracting equations $(1)$ and $(2)$ gives

$4d = 4$ $\implies$ $d = 1$

Substituting $d = 1$ in equation $(1)$ gives

$a + 5 = 3$ $\implies$ $a = -2$

Algebra - Arithmetic Progressions

Find the sum of $20$ terms of an arithmetic progression (A.P) if its first term is $2$ and the seventh term is $20$.

First term of A.P $= a_1 = 2$ (Given)

Let the common difference of A.P $= d$

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

$\therefore \;$ $7^{th}$ term of A.P $= a_7 = a_1 + 6d = 20$ (Given)

i.e. $\;$ $2 + 6d = 20$

i.e. $\;$ $6 d = 18$ $\implies$ $d = 3$

Sum of $n$ terms of A.P $= S_n = \dfrac{n \left[2a_1 + \left(n - 1\right) d\right]}{2}$

$\therefore \;$ Sum of $20$ terms of A.P $= S_{20} = \dfrac{20 \times \left[4 + \left(20 - 1\right) \times 3\right]}{2}$

i.e. $\;$ $S_{20} = 10 \times 61 = 610$

Algebra - Arithmetic Progressions

Find the first term $a_1$ and the common difference $d$ of the arithmetic progression (A.P) in which
$a_2 + a_5 - a_3 = 10$, $\;\;$ $a_2 + a_9 = 17$

First term of A.P $= a_1$, $\;$ Common difference of A.P $= d$

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

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

$3^{rd}$ term of A.P $= a_3 = a_1 + 2d$

$5^{th}$ term of A.P $= a_5 = a_1 + 4d$

$9^{th}$ term of A.P $= a_9 = a_1 + 8d$

Given: $\;$ $a_2 + a_5 - a_3 = 10$

i.e. $\;$ $a_1 + d + a_1 + 4d - a_1 - 2d = 10$

i.e. $\;$ $a_1 + 3d = 10$ $\;\;\; \cdots \; (1)$

and $\;$ $a_2 + a_9 = 17$

i.e. $\;$ $a_1 + d + a_1 + 8d = 17$

i.e. $\;$ $2a_1 + 9d = 17$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously, we have

Equation $(1) \times 2 \implies$ $2a_1 + 6d = 20$ $\;\;\; \cdots \; (3)$

Equation $(2) \times 1 \implies$ $2a_1 + 9d = 17$ $\;\;\; \cdots \; (4)$

Subtracting equations $3$ and $4$ gives

$-3 d = 3$ $\implies$ $d = -1$

Substitute $d = -1$ in equation $(1)$ to get

$a_1 - 3 = 10$ $\implies$ $a_1 = 13$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\cos x + 2 \cos 2x = 1$

Given equation: $\;\;\;$ $\cos x + 2 \cos 2x = 1$

i.e. $\;$ $\cos x + 2 \left(2 \cos^2 x -1\right) = 1$

i.e. $\;$ $\cos x + 4 \cos^2 x - 2 = 1$

i.e. $\;$ $4 \cos^2 x + \cos x - 3 = 0$

i.e. $\;$ $4 \cos^2 x + 4 \cos x - 3 \cos x - 3 = 0$

i.e. $\;$ $4 \cos x \left(\cos x + 1\right) - 3 \left(\cos x + 1\right) = 0$

i.e. $\;$ $\left(4 \cos x - 3\right) \left(\cos x + 1\right) = 0$

i.e. $\;$ $4 \cos x - 3 = 0$ $\;$ or $\;$ $\cos x + 1 = 0$

Case 1: $\;$ $4 \cos x - 3 = 0$

i.e. $\;$ $\cos x = \dfrac{3}{4}$

i.e. $\;$ $x = 2 n \pi \pm \cos^{-1} \left(\dfrac{3}{4}\right), \;\;\; n \in Z$

Case 2: $\;$ $\cos x + 1 = 0 = 0$

i.e. $\;$ $\cos x = -1 = \cos \left(\pi\right)$

i.e. $\;$ $x = 2 m \pi + \pi, \;\;\; m \in Z$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $6 \cos^2 x + 5 \sin x - 7 = 0$

Given equation: $\;\;\;$ $6 \cos^2 x + 5 \sin x - 7 = 0$

i.e. $\;$ $6 \left(1 - \sin^2 x\right) + 5 \sin x - 7 = 0$

i.e. $\;$ $6 - 6 \sin^2 x + 5 \sin x - 7 = 0$

i.e. $\;$ $6 \sin^2 x - 5 \sin x + 1 = 0$

i.e. $\;$ $6 \sin^2 x - 3 \sin x - 2 \sin x + 1 = 0$

i.e. $\;$ $3 \sin x \left(2 \sin x - 1\right) - \left(2 \sin x - 1\right) = 0$

i.e. $\;$ $\left(3 \sin x - 1\right) \left(2 \sin x - 1\right) = 0$

i.e. $\;$ $3 \sin x - 1 = 0$ $\;$ or $\;$ $2 \sin x - 1 = 0$

Case 1: $\;$ $3 \sin x - 1 = 0$

i.e. $\;$ $\sin x = \dfrac{1}{3}$

i.e. $\;$ $x = n \pi + \left(-1\right)^n \sin^{-1} \left(\dfrac{1}{3}\right), \;\;\; n \in Z$

Case 2: $\;$ $2 \sin x - 1 = 0$

i.e. $\;$ $\sin x = \dfrac{1}{2} = \sin \left(\dfrac{\pi}{6}\right)$

i.e. $\;$ $x = m \pi + \left(-1\right)^m \times \dfrac{\pi}{6}, \;\;\; m \in Z$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $1 + \cos x + \cos 2 x = 0$

Given equation: $\;\;\;$ $1 + \cos x + \cos 2 x = 0$

i.e. $\;$ $1 + \cos x + 2 \cos^2 x - 1 = 0$

i.e. $\;$ $2 \cos^2 x + \cos x = 0$

i.e. $\;$ $\cos x \left(2 \cos x + 1\right) = 0$

i.e. $\;$ $\cos x = 0$ $\;$ or $\;$ $2 \cos x + 1 = 0$

Case 1: $\;$ $\cos x = 0 = \cos \left(\dfrac{\pi}{2}\right)$

i.e. $\;$ $x = \left(2n + 1\right) \times \dfrac{\pi}{2}, \;\;\; n \in Z$

Case 2: $\;$ $2 \cos x + 1 = 0$

i.e. $\;$ $\cos x = \dfrac{-1}{2} = \cos \left(\pi - \dfrac{\pi}{3}\right)$

i.e. $\;$ $\cos x = \cos \left(\dfrac{2 \pi}{3}\right)$

i.e. $\;$ $x = 2 m \pi \pm \dfrac{2 \pi}{3}, \;\;\; m \in Z$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\cos 2x + 3 \sin x = 2$

Given equation: $\;\;\;$ $\cos 2x + 3 \sin x = 2$

i.e. $\;$ $1 - 2 \sin^2 x + 3 \sin x = 2$

i.e. $\;$ $2 \sin^2 x - 3 \sin x + 1 = 0$

i.e. $\;$ $2 \sin^2 x - 2 \sin x - \sin x + 1 = 0$

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

i.e. $\;$ $2 \sin x - 1 = 0$ $\;$ or $\;$ $\sin x - 1 = 0$

Case 1: $\;$ $2 \sin x - 1 = 0$

i.e. $\;$ $\sin x = \dfrac{1}{2} = \sin \left(\dfrac{\pi}{6}\right)$

i.e. $\;$ $x = n \pi + \left(-1\right)^n \times \dfrac{\pi}{6}, \;\;\; n \in Z$

Case 2: $\;$ $\sin x - 1 = 0$

i.e. $\;$ $\sin x = 1 = \sin \left(\dfrac{\pi}{2}\right)$

i.e. $\;$ $x = 2 m \pi + \dfrac{\pi}{2}, \;\;\; m \in Z$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\sqrt{3} \sin x - \tan x + \tan x \sin x - \sqrt{3} = 0$

Given equation: $\;\;\;$ $\sqrt{3} \sin x - \tan x + \tan x \sin x - \sqrt{3} = 0$

i.e. $\;$ $\sqrt{3} \left(\sin x - 1\right) + \tan x \left(\sin x - 1\right) = 0$

i.e. $\;$ $\left(\tan x + \sqrt{3}\right) \left(\sin x - 1 = 0\right)$

i.e. $\;$ $\tan x + \sqrt{3} = 0$ $\;$ or $\;$ $\sin x - 1 = 0$

Case 1: $\;$ $\tan x + 3 = 0$

i.e. $\;$ $\tan x = - \sqrt{3}$

i.e. $\;$ $x = n \pi + \tan^{-1} \left(-\sqrt{3}\right)$

i.e. $\;$ $x = n \pi - \tan^{-1} \left(\sqrt{3}\right)$

i.e. $\;$ $x = n \pi - \dfrac{\pi}{3}, \;\;\; n \in Z$

Case 2: $\;$ $\sin x - 1 = 0$

i.e. $\;$ $\sin x = 1 = \sin \left(\dfrac{\pi}{2}\right)$

i.e. $\;$ $x = 2 m \pi + \dfrac{\pi}{2}, \;\;\; m \in Z$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\tan 2x \sin x + \sqrt{3} \left(\sin x - \sqrt{3} \tan 2x\right) = 3 \sqrt{3}$

Given equation: $\;\;\;$ $\tan 2x \sin x + \sqrt{3} \left(\sin x - \sqrt{3} \tan 2x\right) = 3 \sqrt{3}$

i.e. $\;$ $\tan 2x \sin x + \sqrt{3} \sin x - 3 \tan 2x - 3 \sqrt{3} = 0$

i.e. $\;$ $\tan 2x \left(\sin x - 3\right) + \sqrt{3} \left(\sin x - 3\right) = 0$

i.e. $\;$ $\left(\sin x - 3\right) \left(\tan 2x + \sqrt{3}\right) = 0$

$\implies$ $\sin x - 3 = 0$ $\;$ or $\tan 2x + \sqrt{3} = 0$

Case 1:

$\sin x - 3 = 0$

i.e. $\;$ $\sin x = 3$

But sine of any angle cannot be greater than $1$.

$\therefore \;$ $\sin x = 3$ does not give a valid solution.

Case 2:

$\tan 2x + \sqrt{3} = 0$

i.e. $\;$ $\tan 2x = - \sqrt{3}$

i.e. $\;$ $2x = n \pi + \tan^{-1} \left(- \sqrt{3}\right)$

i.e. $\;$ $2x = n \pi - \tan^{-1} \left(\sqrt{3}\right)$

i.e. $\;$ $2x = n \pi - \dfrac{\pi}{3}$

$\therefore \;$ $x = \dfrac{n \pi}{2} - \dfrac{\pi}{6}, \;\;\; n \in Z$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\tan^3 x - 1 + \dfrac{1}{\cos^2 x} - 3 \cot \left(\dfrac{\pi}{2} - x\right) = 3$

Given equation: $\;\;\;$ $\tan^3 x - 1 + \dfrac{1}{\cos^2 x} - 3 \cot \left(\dfrac{\pi}{2} - x\right) = 3$

i.e. $\;$ $\tan^3 x + \left(-1 + \sec^2 x\right) - 3 \tan x - 3 = 0$

i.e. $\;$ $\tan^3 x + \tan^2 x - 3 \tan x - 3 = 0$

i.e. $\;$ $\tan^2 x \left(1 + \tan x\right) - 3 \left(1 + \tan x\right) = 0$

i.e. $\;$ $\left(1 + \tan x\right) \left(\tan^2 x - 3\right) = 0$

$\implies$ $1 + \tan x = 0$ $\;$ or $\;$ $\tan^2 x - 3 = 0$

Case 1:

$1 + \tan x = 0$ $\implies$ $\tan x = -1$

$\begin{aligned} i.e. \; x & = n \pi + \tan^{-1} \left(-1\right) \\\\ & = n \pi - \tan^{-1} \left(1\right) \\\\ & = n \pi - \dfrac{\pi}{4}, \;\; n \in I \end{aligned}$

Case 2:

$\tan^2 x - 3 = 0$ $\implies$ $\tan x = \pm \sqrt{3}$

When $\;$ $\tan x = \sqrt{3} = \tan \left(\dfrac{\pi}{3}\right)$

$\implies$ $x = m \pi + \dfrac{\pi}{3}, \;\; m \in I$

When $\;$ $\tan x = - \sqrt{3}$

$\begin{aligned} i.e. \; x & = k \pi + \tan^{-1} \left(-\sqrt{3}\right) \\\\ & = k \pi - \tan^{-1} \left(\sqrt{3}\right) \\\\ & = k \pi - \dfrac{\pi}{3}, \;\; k \in I \end{aligned}$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $4 \cos^3 x - 4 \cos^2 x - \cos \left(\pi + x\right) - 1 = 0$

Given equation: $\;\;\;$ $4 \cos^3 x - 4 \cos^2 x - \cos \left(\pi + x\right) - 1 = 0$

i.e. $\;$ $4 \cos^2 x \left(\cos x - 1\right) - \left(- \cos x\right) - 1 = 0$

i.e. $\;$ $4 \cos^2 x \left(\cos x - 1\right) + \cos x - 1 = 0$

i.e. $\;$ $\left(4 \cos^2 x + 1\right) \left(\cos x - 1\right) = 0$

$\implies$ $4 \cos^2 x + 1 = 0$ $\;$ or $\;$ $\cos x - 1 = 0$

When $\;$ $4 \cos^2 x + 1 = 0$

i.e. $\;$ $\cos^2 x = \dfrac{-1}{4}$

But square of any number cannot be negative.

$\therefore \;$ $\cos^2 x = \dfrac{-1}{4}$ $\;$ will not give a valid solution.

When $\;$ $\cos x - 1 = 0$

i.e. $\;$ $\cos x = 1 = \cos 0$

$\implies$ $x = 2 n \pi, \;\; n \in I$