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Oduvanchick [21]

3 years ago

8

3 (bx - 2ab) = b (x - 7a) + 3ab solve for x

2 answers:

geniusboy [140]3 years ago

4 0

<span> 3(bx - 2ab) = b(x - 7a) + 3ab
<=> </span><span>3bx - 6ab = bx - 7ab + 3ab
<=> 3bx - bx = 6ab - 7ab + 3ab
<=> 2bx = 2ab
<=> x = a</span>

chubhunter [2.5K]3 years ago

3 0

Answer:

The value of x = a.

Step-by-step explanation:

Consider the provided equation.

3 (bx - 2ab) = b (x - 7a) + 3ab

Use the distributive property: a(b+c)=ab+ac

3bx - 6ab = bx - 7ab + 3ab

Now add the like variable.

3bx - 6ab = bx - 4ab

Subtract bx and add 6ab on both sides.

3bx - 6ab - bx + 6ab = bx - bx - 4ab + 6ab

2bx = 2ab

Divide both the side by 2b.

x = a

Hence, the value of x = a.

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The 4th and the last terms of an A.P. are 11 and 89 respectively. If there are 30 terms in the A.P., find the A.P. and its 23rd

Natali5045456 [20]

$\underline{\underline{\large\bf{Given:-}}}$

$\red{\leadsto}\:$ $\textsf{}$ $\sf Number \: of \:terms \: in \: A.P,n = 30$

$\red{\leadsto}\:$ $\textsf{}$ $\sf Fourth \: term ,a_4 = 11$

$\red{\leadsto}\:$ $\textsf{}$ $\sf last\:term, a_{30} = 89$

$\underline{\underline{\large\bf{To Find:-}}}$

$\orange{\leadsto}\:$ $\textsf{ }$ $\sf The \: A.P.$

$\orange{\leadsto}\:$ $\textsf{ }$ $\sf 23rd\: term, a_{23}$

$\\$

$\underline{\underline{\large\bf{Solution:-}}}\\$

The nth term of A.P is determined by the formula-

$\green{ \underline { \boxed{ \sf{a_n = a+(n-1)d}}}}$

where

$\sf a = first \:term$
$\sf a_n = nth \: term$
$\sf n = number \:of \:terms$
$\sf d = common \: difference$

Since ,

$\sf a_4 = 11$

$\longrightarrow$ $\sf a+(4-1) d= 11$

$\longrightarrow$ $\sf a+3d= 11\_\_\_(1)$

$\sf a_{30}= 89$

$\longrightarrow$ $\sf a+(30-1)d=89$

$\longrightarrow$ $\sf a+29d= 89\_\_\_(2)$

<u>Subtracting equation (1) from equation(2)</u>

$\begin{gathered}\\\implies\quad \sf a+29d-(a+3d) = 89-11 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a+29d-a-3d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a-a+29d-3d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf 26d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf d = \frac{78}{26} \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf d = 3 \\\end{gathered}$

Putting the value of d in equation (1) -

$\begin{gathered}\\\implies\quad \sf a+3(3) = 11 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a = 11-9 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a = 2 \\\end{gathered}$

First term of A.P, a = 2

Second term of A.P., $\sf a_2= 2+(2-1)\times 3$

$\quad\quad\quad\sf =2+3$

$\quad\quad\quad\sf =5$

Third term of A.P., $\sf a_3= 2+(3-1)\times 3$

$\quad\quad\quad\sf =2+6$

$\quad\quad\quad\sf =8$

$\longrightarrow$ Thus , The A.P is 2,5,8,. . . . . .

<u>Now,</u>

$\begin{gathered}\\\implies\quad \sf a_n = a+(n-1)d \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23 }= 2+(23-1)\times 3 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 2+22 \times 3 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 2+66 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 68 \\\end{gathered}$

$\longrightarrow$ Thus , 23rd term is 68.

3 0

3 years ago