Asked by Da silva
The second term of a geometric progression is 13 more than the first term given that the common ratio is half the first term. Find the 3rd term of the G.P.
Answers
Answered by
mathhelper
ar - a = 13 and r = a/2
a(a/2) - a = 13
a^2 - 2a - 26 = 0
a = 1 ± 3√3
if a = 1 + 3√3, then r = (1+3√3)/2
and term(3) = ar^2 = (1+3√3) (1+ 3√3)^2 / 4
if a = 1 - 3√3, then r = (1-3√3)/2
and term(3) = ar^2 = (1-3√3) (1- 3√3)^2 / 4 , I will let you evaluate those
but checking
terms are : 1+3√3, (1+3√3)^2 / 2
difference between 2nd and first = (1+3√3)^2 / 2 - (1+3√3)
= (1 + 6√3 + 27)/2 - 1 - 3√3
= (28 + 6√3)/2 -1 - 3√3
= 14 + 3√3 - 1 - 3√3
= 13 , as required
a(a/2) - a = 13
a^2 - 2a - 26 = 0
a = 1 ± 3√3
if a = 1 + 3√3, then r = (1+3√3)/2
and term(3) = ar^2 = (1+3√3) (1+ 3√3)^2 / 4
if a = 1 - 3√3, then r = (1-3√3)/2
and term(3) = ar^2 = (1-3√3) (1- 3√3)^2 / 4 , I will let you evaluate those
but checking
terms are : 1+3√3, (1+3√3)^2 / 2
difference between 2nd and first = (1+3√3)^2 / 2 - (1+3√3)
= (1 + 6√3 + 27)/2 - 1 - 3√3
= (28 + 6√3)/2 -1 - 3√3
= 14 + 3√3 - 1 - 3√3
= 13 , as required
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