Question

please reply if anyone knows

Answer 1

k+1,3k,4k+2,......-->A.P

If k=1,

A.P:2,3,6,....--->Not an AP

If k=2,

A.P:3,6,10,...--->Not an AP

If k=3,

A.P:4,9,14,....--->AP with common difference 5.

If k=4,

A.P:5,12,18--->Not an AP.

Hence,

k=3.

Answer 2

As we know that for an AP, the middle term will be equal to the avg of the sum of it's neighbouring terms...

ie

If 2, 4, 6, 8, 10 are in AP, then

4=(2+6)/2

6=(4+8)/2

8=(6+10)/2

So applying the same concept here in this question:

it is given that k+1, 3k & 4k+2 are in AP

So

3k=[(k+1)+(4k+2)]/2

3k=(5k+3)/2
6k=5k+3
k=3

I hope it helped u...

Pls Brainliest it if it helped u☺️☺️☺️♥️♥️