Class 10

Math

All topics

Arithmetic Progressions

Determine k so that $(3k−2),(4k−6)$ and $(k+2)$ are three consecutive terms of an AP.

Given: $(3k−2),(4k−6)$ and $(k+2)$ are three consecutive terms of an AP.

So, common difference between two consecutive terms will be same

$(4k−6)−(3k−2)=(k+2)−(4k−6)$

$2(4k−6)=(k+2)+(3k−2)$

$8k−12=4k+0$

$8k−4k=0+12$

or $k=3$.