Question

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




Pls​

Answer 1

Answer:

$\bigstar{\bold{Value\:of\:x=\dfrac{15}{2} }}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• First term (a) = 8x + 4
• Second term (b) = 6x - 2
• Third term (c) = 2x + 7

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The value of x

$\Large{\underline{\underline{\bf{Solution:}}}}$

➛ Here we have to find the value of x so that the 3 terms are in A.P

➛ We know that if 3 terms are in A.P,

    c - b = b - a

    a + c = 2b

    b = (a + c)/2

➛ Substituting the data,

    6x - 2 = (8x + 4 + 2x + 7)/2

➛ Simplifying,

   12x - 4 = 8x + 4 + 2x + 7

   12x - 4 = 10x + 11

   12x - 10x = 11 + 4

   2x = 15

      x = 15/2

➛ Hence the value of x is 15/2

   $\boxed{\bold{Value\:of\:x=\dfrac{15}{2} }}$

$\Large{\underline{\underline{\bf{Verification:}}}}$

➠ The first term of the A.P = 8x + 4

➠ Substituting the data,

    First term = 8 × 15/2 + 4

    First term = 64

➠ Second term = 6x - 2

    Second term =  6 × 15/2 - 2

    Second term = 43

➠ Third term = 2x + 7

    Third tem =  2 × 15/2 + 7

    Third term = 22

➠ If the 3 term are in A.P,

    c - b = b - a

➠ Substitute the data,

    22 - 43 = 43 - 64

    -21 = -21

➠ Hence the common difference of the terms is same

➠ Hence they are in A.P

➠ Hence verified.

$\Large{\underline{\underline{\bf{Notes:}}}}$

⇢ The common difference of an A.p is given by,

    d = a₂ - a₁

    $\sf{d=\dfrac{a_m-a_n}{m-n} }$