Geometric progressions

[janekela]

I got a number of sums in class but I dont even know how to start this one. Could someone please tell me if the first one is correct and help me to do the other?

In a geometric progression (that is, a geometric sequence), the first term exceeds the third by 72 and the sum of the second and third term is 36. Find the first term.

Please help. Thank you!

 

[soroban]

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]: \(\displaystyle \L\,a\:=\:\frac{73}{1\,-\,\left(\frac{1}{3}\right)^2} \:=\:\frac{72}{\frac{8}{9}}\:=\:81\)