Hello, janekela!
In a geometric progression, the first term exceeds the third by 72
and the sum of the second and third term is 36.
Find the first term.
Let \(\displaystyle a\) = first term. and \(\displaystyle r\) = common ratio.
The terms of the progression are: \(\displaystyle \,a,\;ar,\;ar^2,\;ar^3,\;ar^4,\;...\)
"The first term exceeds the third by 72": \(\displaystyle \;a\:=\:ar^2\,+\,72\;\)
[1]
"The sum of the second and third term is 36": \(\displaystyle \;ar\,+\,ar^2\:=\:36\;\)
[2]
From
[1], we have: \(\displaystyle \,a\,-\,ar^2\:=\:72\;\;\Rightarrow\;\;a(1\,-\,r^2)\:=\:72\;\;\Rightarrow\;\;\L a\:=\:\frac{72}{1\,-\,r^2}\;\)
[3]
From
[2], we have: \(\displaystyle \,ar(1\,+\,r)\:=\:36\;\;\Rightarrow\;\;\L a\:=\:\frac{36}{r(1\,+\,r)}\;\)
[4]
Equate
[3] and
[4]: \(\displaystyle \L\,\frac{72}{1\,-\,r^2}\:=\:\frac{36}{r(1\,+\,r)}\;\;\Rightarrow\;\;\)\(\displaystyle 72r(1\,+\,r)\:=\:36(1\,-\,r)(1\,+\,r)\)
Simplify: \(\displaystyle \,2r\:=\:1\,-\,r\;\;\Rightarrow\;\;3r\,=\,1\;\;\Rightarrow\;\;r\,=\,\frac{1}{3}\)
Substitute into
[3]: \(\displaystyle \L\,a\:=\:\frac{73}{1\,-\,\left(\frac{1}{3}\right)^2} \:=\:\frac{72}{\frac{8}{9}}\:=\:81\)
