Find the value of $x$ for which $(8 x + 4), (6 x - 2)$ and $(2 x + 7)$ are in AP.

\frac{13}{2}

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\frac{11}{3}

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\frac{17}{3}

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\frac{15}{2}

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Solution

The correct option is D $\frac{15}{2}$
Since $(8 x + 4), (6 x - 2), (2 x + 7)$ are in AP. Therefore,
$2 \times (6 x - 2) = (8 x + 4) + (2 x + 7)$
$\Rightarrow 12 x - 4 = 10 x + 11$
$\Rightarrow 2 x = 15$
$\Rightarrow x = \frac{15}{2}$

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Similar questions

Q. Find the value of

x

for which

(8 x + 4), (6 x - 2)

and

(2 x + 7)

are in A.P.

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

Q. Find the value of x for which (8x + 4), (6x - 2) and (2x + 7) are in A.P.

Q. if

x = \frac{\sqrt{3} + 1}{2}

,find the value of

4 x^{3} + 2 x^{2} - 8 x + 7

Q. Find the value of
(i) 1 + 3 + 5 = _____
(ii) 1 + 3 + 5 + 7 + 9 +11 = ________
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 =_____

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Find the value of x for which (8x+4),(6x−2) and (2x+7) are in AP.

The correct option is D 152Since(8x+4),(6x−2),(2x+7) are in AP. Therefore,2×(6x−2)=(8x+4)+(2x+7)⇒12x−4=10x+11⇒2x=15⇒x=152

Find the value of $x$ for which $(8 x + 4), (6 x - 2)$ and $(2 x + 7)$ are in AP.

The correct option is D $\frac{15}{2}$
Since $(8 x + 4), (6 x - 2), (2 x + 7)$ are in AP. Therefore,
$2 \times (6 x - 2) = (8 x + 4) + (2 x + 7)$
$\Rightarrow 12 x - 4 = 10 x + 11$
$\Rightarrow 2 x = 15$
$\Rightarrow x = \frac{15}{2}$