Question

If K +1, 3K and 4K +2 are in arithemetic progression than K =​

Answer 1

Given

We have given that k+1,3k & 4k+2 are in ap

To Find

We have to find the value of k

$\sf\huge {\underline{\underline{{Solution}}}}$

Since, all given terms are in ap then it means their common difference must be same .

e.g., a₂-a₁= a₃-a₂

Here, first term (a₁) is k+1

second term (a₂) is 3k

and term (a₃) is 4k+2

Common difference (d)

d for the 1st & 2nd term = a₂-a₁

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=3k-(k+1)= 2k-1

d2 =for the 2nd & 3rd term = a₃-a₂

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=4k+2-(3k)= k +2

Now, equating both d

d1= d2

=> 2k-1= k+2

=> 2k-k= 2+1

=> k = 3

=> k= 3

Therefore ,the values are : k+1= 4

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=>3k= 9

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$=>4k+2= 14

Hence,the ap are : 4,9,14 ..