Question

2. For
what value of t will 2k+1 , 3k+3 , 5k-1
consecutive
a are the
terms of
an A.P.​

Answer 1

Given:

• 2k + 1 , 3k + 3 and 5k - 1 are in AP.

To find :

• The value of k .

Solution:

• As we know that , in AP common difference is same :

$\star \boxed{\bold{t_2 - t_1 = t_3 - t_2} }$

Here:

$\star \: t_1 = \: 2k + 1 \\ \\ \star \: t_2 = 3k + 3 \\ \\ \star \:t_3 = 5k - 1$

$\implies \: 3k + 3 - (2k + 1) = 5k - 1 - (3k + 3) \\ \\ \implies \: 3k + 3 - 2k - 1 = 5k - 1 - 3k - 3 \\ \\ \implies \: k + 2 = 2k - 4 \\ \\ \implies \: 2k - k = 2 + 4 \\ \\ \implies \: k = 6$

Hence , the value of k is 6.

Answer 2

In AP

second - first = thirs - second

=> 3k+3 - 2k -1 = 5k-1 -3k-3

=> k+2 = 2k-4

=> k = 6 .