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
please answer fast

Answer 1

Answer:

$k = 1 \\ \\ (b) \: \: is \: correct$

Step-by-step explanation:

$2(4k + 2) = 3k + (6k + 3) \\ 8k + 4 = 9k + 3 \\ 1 = k \\ or \: \: \: k = 1 \\ \\ \\ therefore \: \: \: option \: (b) \: is \: correct$

Answer 2

Answer:

hope helps u

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Step-by-step explanation:

S O L U T I O N :

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

We have,

a1 = 3k

a2 = (4k + 2)

a3 = (6k + 3)

A/q

By common difference method :

Thus,

The value of k will be 1 .

Option (b)