Solve the following mathematical problems:

Let, second term = a 

Common difference = d

First term will be=a-d 

And third term will be = a+d 

according to question, 

$$\begin{gathered} a-d+a+a+d=30 \\ \Rightarrow 3a=30 \\ \Rightarrow a=\frac{30}{3} \\ \therefore a=10 \end{gathered}$$

So, second term = 10 

First term = (10-d) 

and third = (10+d) 

According to question , 

$$\begin{gathered} {\left(10-d\right)}^2+{\left(10\right)}^2+{\left(10+d\right)}^2=318 \\ \Rightarrow {\left(10\right)}^2-2\times 10\times d+d^2+100+{\left(10\right)}^2+2\times 10\times d+d^2=318 \\ \Rightarrow 100-20d+d^2+100+100+20d+d^2=318 \\ \Rightarrow 2d^2+300=318 \\ \Rightarrow 2d^2=18 \\ \therefore d=\sqrt{9}=3 \end{gathered}$$

so, 1^st term  = 10-3=7; 2^nd term= 10 & 3^rd term = 10+3=13 

ans. the number is 7,10,13 

Here, the necessary rule is

 Total = English+ Bangla - Both + None

 50 = 35+ Bangla - 25+0 

Bangla = 40 

So total 40 speak Bangla 

Speak only dot Bangla = 40 - 25 = 15

ans. Only 15 people speak in Bangla.

Let, x be the perpendicular drawn from the vertex of the triangle 

We know, 

$$\begin{gathered} \sin 45°=\frac{Perpendicular}{Hypotenuse} \\ \Rightarrow \frac{1}{\sqrt{2}}=\frac{x}{10} \\ \therefore x=\frac{10}{\sqrt{2}} \end{gathered}$$

Now, we know area of triangle

$=\frac{1}{2}\times base\times height=\frac{1}{2}\times 10\times \frac{10}{\sqrt{2}}=\frac{50}{\sqrt{2}}=\frac{25\times \sqrt{2}\times \sqrt{2}}{\sqrt{2}}=25\sqrt{2}$ ans.

$$\begin{gathered} we know, {\left(x-y\right)}^3={\left(x\right)}^3-3x^{2}y+3x{y}^2-y^3 \\ Now, 64x^3-9a{x}^2+108x-b \\ ={\left(4x\right)}^3-3\times {\left(4x\right)}^2\times 3+3\times 4x\times {\left(3\right)}^2-{\left(3\right)}^3 \\ ={\left(4x-3\right)}^3 \\ In here, in the 2^{nd} term \\ 3\times {\left(4x\right)}^2\times 3=3\times {\left(4\right)}^2\times x^2\times 3 \\ So,in here 4^2=16=a \\ In the 3^{rd} term= 3\times 4x\times 3^2 \left[3x{y}^2\right] \\ And in 4^{th }terrm= 3^3 = b \\ So, b=3^3=27 \\ \mathit{a}\mathit{n}\mathit{s}\mathbf{.}\mathbf{ }\mathit{a}\mathbf{=}\mathbf{16}\mathbf{ }\mathbf{\&}\mathbf{ }\mathit{b}\mathbf{=}\mathbf{27}\mathbf{ } \end{gathered}$$

Let, the price of one table is x Tk.

 And the price of one chair is y Tk. 

According to the question

 3x + 5y =2,000....................(i) 

5x + 7y = 3, 200 ……..(ii) 

Now, we multiply equation (i) by 5 & (ii) by 3 & subtract equation (ii) from equation (i) 

$$\begin{gathered} 15x + 25y = 10000 \\ 15x+21y= 9.600 \\ 4y = 400 \\ \Rightarrow y = \frac{400}{4} \\ \Rightarrow y = 100 \end{gathered}$$

Putting the value of y in equation (i), we get 

 $$\begin{gathered} 3x + 5y = 2000 \\ \Rightarrow 3x + (5 * 100) = 2000 \\ \Rightarrow 3x + 500 = 2000 \\ \Rightarrow 3x = 2000 - 500 \\ \Rightarrow 3x = 1,500 \\ \therefore x=\frac{1500}{3}=500 \\ x=500 \end{gathered}$$

ans. The price of 1 table is 500 Tk. and 1 chair is 100 Tk.

Committee members will be 3 

Available men = 7 & Available women = 5

 So, the combinations can be

So, total combinations = 35 + 105 + 70 + 10 = 220 ways. ans.

Total business time = 8+7= 15 months. 

A's time weighted investment =$[3,000 \times 8+ (3,000+ 2,500) \times 7] = 24,000+ 38,500 = 62,500 Tk.$ 

 B's time weighted investment = $4000\times 15 = 60000 Tk .$

So, investment ratio = $62,500 : 60,000 = 25 : 24$

Now, sum of their invest t = 25 + 24 = 49  [Dividing by 2,500]

So A get profit =$\left(980\times \frac{25}{49}\right)=500$ Tk & B get profit =$\left(980 \times \frac{24}{49} \right)= 480$ Tk

 ans. The share of profit is 500 & 480 Tk.

$$\begin{gathered} a^2+\frac{1}{a^2}+2-2a-\frac{2}{a} \\ \Rightarrow {\left(a+\frac{1}{a}\right)}^2-2\times a\times \frac{1}{a}+2-2a-\frac{2}{a} \\ \Rightarrow {\left(a+\frac{1}{a}\right)}^2-2+2-2a-\frac{2}{a} \\ \Rightarrow {\left(a+\frac{1}{a}\right)}^2-2a-\frac{2}{a} \\ \Rightarrow {\left(a+\frac{1}{a}\right)}^2-2\left(a+\frac{1}{a}\right) \\ \Rightarrow \left(a+\frac{1}{a}\right)\left(a+\frac{1}{a}\right)-2\left(a+\frac{1}{a}\right) \\ \Rightarrow \left(a+\frac{1}{a}\right)\left(a+\frac{1}{a}-2\right)\mathit{a}\mathit{n}\mathit{s}\mathbf{.} \end{gathered}$$