Class 10

Math

All topics

Arithmetic Progressions

The sum of first three terms of an AP is $48$. If the product of first and second terms exceeds $4$ times the third term by $12$. Find the AP.

Let $a−d,a,a+d$ are the three terms of AP

So,

Sum$=a−d+a+a+d=48$

$3a=48$

or $a=16$

And,

$a(a−d)=4(a+d)+12$

$16(16−d)=4(16+d)+12$

$256−16d=64+4d+12=4d+76$

$256−76=4d+16d$

$180=20d$

$d=9$

Which implies:

Numbers are: $(7,16,25)$.