Question

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

Answer 1

Question :–

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

Answer :–

$\tt Value \: of \: x = 7.5$

Given :–

1. 1st term = 8x + 4
2. 2nd term = 6x - 2
3. 3rd term = 2x + 7

To Find :–

• Value of x.

Solution :–

Common difference (d) of first two terms :-

$\implies \sf d = a_{2} - a_{1}$

$\implies \sf (6x - 2) - (8x + 4)$

$\implies \sf 6x - 2 - 8x - 4$

$\implies \sf 6x - 8x - 2 - 4$

$\implies \sf - 2x - 6$

Common difference (d) of last two terms :-

$\implies \sf d = a_{3} - a_{2}$

$\implies \sf (2x + 7) - (6x - 2)$

$\implies \sf 2x + 7 - 6x + 2$

$\implies \sf 2x - 6x + 7 + 2$

$\implies \sf - 4x + 9$

Common difference of an A.P. :-

$\implies \sf d = a_{2} - a_{1} = a_{3} - a_{2}$

$\implies \sf ( - 2x - 6) = ( - 4x + 9)$

$\implies \sf - 2x - 6 = - 4x + 9$

$\implies \sf - 2x + 4x = 9 + 6$

$\implies \sf 2x = 15$

$\implies \sf x = \dfrac{15}{2}$

$\implies \sf x = 7.5$

Hence,

Value of $\tt x = 7.5$

Answer 2

Given :

• First term = 8x + 4
• Second term = 6x - 2
• Third term = 2x + 7

To Find :

• Find the value of x.

Solution :

$\dashrightarrow\:\sf d = a_2 - a_1 \\ \\ \dashrightarrow\:\sf d = \left(6x - 2 \right) - \left(8x + 4 \right) \\ \\ \dashrightarrow\:\sf d = 6x - 8x - 2 - 4 \\ \\ \red\dashrightarrow\: \red{\sf d = - 2x - 6----(1)}$

Also,

$\dashrightarrow\:\sf d = a_3 - a_2 \\ \\\dashrightarrow\: \sf d = \left(2x + 7 \right) - \left(6x - 2 \right) \\ \\ \dashrightarrow\: \sf d = 2x - 6x + 7 + 2 \\ \\\pink\dashrightarrow\: \pink{\sf d = - 4x + 9----(2)}$

Now, equating equation (1) and (2) we get :

$\dashrightarrow\:\sf - 2x - 6 = - 4x + 9 \\ \\ \dashrightarrow\:\sf 4x - 2x = 9 + 6 \\ \\\dashrightarrow\:\sf 2x = 15 \\ \\\dashrightarrow\:\sf x = \dfrac{15}{2} \\ \\\dashrightarrow\:\underline{\boxed{\gray{\sf x = 7.5}}}$

Therefore, the value of x = 7.5 .