if the sum of three numbers in a GP is 7/64 and the product of the extremes is 1 by 1204 then find numbers
Answers
let the 3 nos be a , ar and ar²
given that the sum of no's is 70
there for a(1+r+r²) =70 ....................................................(i)
also given that 4a,5ar and 4ar² in AP
⇒2(5ar)=4a+4ar²
⇒5r=2+2r²
⇒2r²-5r+2=0
r=2 or 1/2
from(i) we get
a=10 at r=2
and a =40 at r=1/2
where a=10 and r=2 , the numbers are 10, 20, ,40
and when a=40 and r =1/2
the numbers are 40,20,10
hence the numbers are 10,20,40
Answer:
Step-by-step explanation:
Apmpman Brainly Challenger
let the 3 nos be a , ar and ar²
given that the sum of no's is 70
there for a(1+r+r²) =70 ....................................................(i)
also given that 4a,5ar and 4ar² in AP
⇒2(5ar)=4a+4ar²
⇒5r=2+2r²
⇒2r²-5r+2=0
r=2 or 1/2
from(i) we get
a=10 at r=2
and a =40 at r=1/2
where a=10 and r=2 , the numbers are 10, 20, ,40
and when a=40 and r =1/2
the numbers are 40,20,10
hence the numbers are 10,20,40