Question

 If 7+k ,22-k, and 5k+5 are in A.P then k is​

Answer 1

Answer:

• k = 3/2

Explanation:

Given A. P.

⇒ (7 + k), (22 - k), and (5k + 5)

We know that difference between the first two terms will be equal to last two terms.

So,

⇒ (22 - k) - (7 + k) = (5k + 5) - (22 - k)

⇒ 22 - k - 7 - k = 5k + 5 - 22 + k

⇒ 29 - 2k = 6k + 17

⇒ 29 - 17 = 6k + 2k

⇒ 12 = 8k

⇒ 12/8 = k

⇒ 3/2 = k

Hence, the value of ‘k’ is 3/2

_____________________

Note :

Difference between every two term in a sequence is always same and thus called common difference.

For example :

An A. P. is 10, 20, 30, 40....

Here common difference is 10

Answer 2

Given :

• 7 + k , 22 - k and 5k + 5 are in A.P

To find :

• K = ?

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━$

$\sf{We \: know \: that, \: The \: difference \: between \: the \: two \: terms \: will \: be \: equal \: to \: last \: two \: terms.}$

$\tt{(22 - k) - (7 + k) =( 5k + 5) -( 22 - k) }$

$\tt{ \qquad : \implies \: 22 - k - 7 - k = 5k + 5 - 22 + k}$

$\tt{ \qquad \: \: \: \: \: \: \: \: : \implies \: 29 - 2k = 6k + 17}$

$\tt{ \qquad \: \: \: \: \: \: \: \: : \implies \: 29 - 17 = 6k + 2k}$

$\tt{ \qquad \: \: \: \: \: \: \: \: : \implies \: 12 = 8k}$

$\tt{ \qquad \: \: \: \: \: \: \: \: : \implies \: k = \cancel \frac{12}{8} }$

${\frak{ \qquad \: \: \: \: \: \: \: \: : \implies \underline{ \boxed {\frak \green{k = \frac{3}{2} }}}}}$