Question

if 2 k + 1,3 K + 3,5 K - 1 forms an ap then what is the value of k​

Answer 1

Answer:k=6

Step-by-step explanation:since 2k+1,3k+3 and 5k-1 are in the form of an AP

    so, (3k+3)-(2k+1)=(5k-1)-(3k+3)        (as the difference is same)

        =>3k+3-2k-1=5k-1-3k-3

        =>k+2=2k-4

        =>2k-k=2+4

        =>k=6

Answer 2

Answer:

k = 6

Step-by-step explanation:

Given A.P. : 2k + 1, 3k + 3, 5k - 1

From the above arithmetic progession, we get the following information :

${\sf{a_1}}$ = 2k + 1

${\sf{a_2}}$ = 3k + 3

${\sf{a_3}}$ = 5k - 1

Now,

Common Difference, d = ${\sf{a_2 - a_1}}$

Also, d = ${\sf{a_3 - a_2}}$

Therefore,

$\Rightarrow{\sf{a_2 - a_1 = a_3 - a_2}}$

Putting known values, we get

$\Rightarrow$ (3k + 3) - (2k + 1) = (5k - 1) - (3k + 3)

$\Rightarrow$ 3k + 3 - 2k - 1 = 5k - 1 - 3k - 3

$\Rightarrow$ 3k - 2k + 3 - 1 = 5k - 3k - 1 - 3

$\Rightarrow$ k + 2 = 2k - 4

$\Rightarrow$ 2k - k = 4 + 2

$\Rightarrow$ k = 6