Question

The sum and product of theee consecutive terms of an AP are 12,18,respectively find 3 terms

Answer 1

Answer:

Assuming the 3 numbers to be consecutive terms of A.P. Then we have these terms as; a1=a−d,a2=a,a3=a+d

⇒a=6 adding them will give us; 3a=18

adding 2,4,11 to corresponding terms will result in GP. i.e

a1=(6−d)+2,

a2=6+4,

a3=(6+d)+11

Now by property of G.P we have

108−d=17+d10

⇒d2+9d−36=0 ⇒(d+12)(d−3)=0 ⇒d−3=0ord+12=0

Thus if d=3 and a=6 the numbers are a1=6−3, a2=6 a3=6+4i.ea1=3;a2=6;a3=9;$

⇒d=3ord=12

Thus if d=3 and a=6 the numbers are a1=6−3,a2=6a3=6+4i.ea1=3;a2=6;a3=9;

and if d=−12anda=6we have a1=18; a2=6; a3=−6

I Hope It Will Help!

^_^