Question

For what value of k ,are the numbers x, 2x+k and 3x+6 ,three consecutive terms of an A.P.​

Answer 1

Answer:

$\sf{The \ value \ of \ k \ is \ 3.}$

Given:

$\sf{Three \ consecutive \ terms \ of \ an \ AP}$

$\sf{are \ x, \ 2x+k \ and \ 3x+6.}$

To find:

$\sf{The \ value \ of \ k.}$

Solution:

$\sf{Here,}$

$\sf{t_{1}=x,}$

$\sf{t_{2}=2x+k,}$

$\sf{t_{3}=3x+6}$

$\boxed{\sf{2\times \ t_{2}=t_{1}+t_{3}}}$

$\sf{\therefore{2(2x+k)=(x)+(3x+6)}}$

$\sf{\therefore{4x+2k=4x+6}}$

$\sf{\therefore{2k=6}}$

$\sf{\therefore{k=\dfrac{6}{2}}}$

$\sf{\therefore{k=3}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ 3.}}}$

Answer 2

$\large{\boxed{\bf{Answer\ :-}}}$

• The value of K is 3

__________________________

Given :-

• Three consecutive terms of an AP as x, (2x+k), (3x+6).

To Find :-

• Value of k

__________________________

$\large{\boxed{\bf{Solution}}}$

• T1 = x

• T2 = 2x + k

• T3 = 3x + 6

★ 2 T2 = T1 + T3

$2(2x + k) = x + (3x + 6)$

$= > 4x + 2k = x + 3x + 6$

$= > 4x + 2k = 4x + 6$

$= > 2k = 6$

$= > k = \frac{6}{2}$

$= > k = 3$

Hence, the value of k is 3 .