Question

If 3k, (4k+2), (6k+3) are three consecutive terms of an A.P , then find the value
of k is (a) 3 (b) 1 (c) - 1 (d) 6 (e) none of these

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

If 3k, (4k +2), (6k+3) are the three consecutive terms of an A.P.

$\underline{\bf{Explanation\::}}$

We have,

• a1 = 3k
• a2 = (4k + 2)
• a3 = (6k + 3)

A/q

By common difference method :

$\longrightarrow\tt{a_2 - a_1 = a_3 - a_2}$

$\longrightarrow\tt{(4k + 2) -3k = (6k+3) - (4k + 2)}$

$\longrightarrow\tt{4k + 2 -3k = 6k+3 - 4k -2}$

$\longrightarrow\tt{k + 2 = 2k+1}$

$\longrightarrow\tt{k-2k=1-2}$

$\longrightarrow\tt{\cancel{-}k=\cancel{-}1}$

$\longrightarrow\bf{k=1}$

Thus,

The value of k will be 1 .

Option (b)