Question

9. For what value of k, will the terms 2k + 1,8 and 3k form an A.P.​

Answer 1

Here is your answer user.

Answer 2

SolutioN :

We have, Terms in AP

• 2k + 1 , 8 and 3k .

☛ Condition.

• AP can be define as When their Common different be same.

Let,

• first term be a
• Common difference be d.
• a1 = 2k + 1.
• a2 = 8.
• a3 = 3k.

A / Q,

$\tt \dagger \: \: \: \: \: \fbox{a_2-a_1=d}$

$\tt \dagger \: \: \: \: \: \fbox{a_3-a_2=d}$

✎ So, We can also write as.

$\tt : \implies a_2-a_1 = a_3 - a_2.$

$\tt : \implies 8-(2k + 1)= 3k - 8.$

$\tt : \implies 16-2k - 1= 3k .$

$\tt : \implies 16 - 1= 5k .$

$\tt : \implies 15= 5k .$

$\tt : \implies k = \dfrac{ \cancel{15}}{ \cancel5}$

$\tt : \implies k = 3.$

♢ Therefore, the value of k is 3.