Question

If a1,a2,a3 are in gp such that a1 + a2 + a3 =13 and (a1)^2 +(a2)^2 +(a3)^2 = 91 then find a and r ​

Answer 1

Step-by-step explanation:

(a1)²+(a2)²+(a3)²=(a1²+a2²+a3²). +2(a1a2+a2a3+a1a3)

13²=91+2(a1a2+a2a3+a1a3)

(a1+a2+a3)=(13²-91)/2

=(169-91)/2

= 78/2

=39

Therefore,if (a1+a2+a3)=13 and

(a1a2+a2a3+a1a3)=39 then

r= (a1a2+a2a3+a1a3)/(a1+a2+a3)

r=39/13

r=3

So assuming a=1 and r=3,the geometries series is 1,3,9

i.e.a1=1

a2=3

a3=9

Answer 2

Answer:

a=9,1 ,r=1/3,3,Sn=13

Step-by-step explanation: