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3 years ago

15

2(6 − 2x) + 3x = 4 + 2(5 + x) what is x

1 answer:

alexira [117]3 years ago

5 0

First let's get rid of the parenthesis:

12 - 4x + 3x = 4 + 10 + 2x

Now group the x's on the left and the numbers on the right

-4x + 3x - 2x = 4 + 10 - 12

...and simplify

-3x = 2

divide by -3 to isolate x

x = -2/3

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Sauron [17]

$\rule{200}4$

Answer : <em>The</em><em> </em><em>required</em><em> </em><em>ratio</em><em> </em><em>is</em><em> </em><em>(</em><em>1</em><em>4</em><em>m</em><em>-</em><em>6</em><em>)</em><em>:</em><em>(</em><em>8</em><em>m</em><em>+</em><em>2</em><em>3</em><em>)</em><em> </em><em>.</em>

$\rule{200}4$

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27} \\$

As , we know that the sum of n terms of an AP is given by ,

$\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\$

Assume that ,

First term of 1st AP = a
First term of 2nd AP = a'
Common difference of 1st AP = d
Common difference of 2nd AP = d'

Using this we have ,

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\$

$\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\$

Now also we know that the nth term of an AP is given by ,

$\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\$

Therefore,

$\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\$

$\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\$

From equation (i) and (iii) ,

$\longrightarrow\sf\small n-1 = 2m-2\\$

$\longrightarrow\sf\small n = 2m -2+1 \\$

$\longrightarrow\sf\small n = 2m -1 \\$

Substitute this value in equation (i) ,

$\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\$

Simplify,

$\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\$

$\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

$\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

From equation (ii) ,

$\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\$

$\rule{200}4$

8 0

2 years ago