Problem- The sum of the first two terms of an infinite geometric series is 18. Also, each term of the series is seven times the sum of all the terms that follow. Find the first term and the common ratio of the series respectively.
My approach- Let $a+ar+ar^2+\dots$ be the series. Then,
$a+ar=18$ and,
$a=7\frac{1}{1-r}-a$,
solving I get $r = \frac{29}{43}$ but
Given answer is $a=16,r=1/8$.
where I'm doing wrong?
Best Answer
You are correct with the first equation $a+ar=18$. This implies that $a\neq 0$. Applying the second condition to the first term, we get $$a=7(ar+ar^2+ar^3+\dots)$$ that is, we get $$a=7[-a+(a+ar+ar^2+ar^3+\dots)]$$ implying that $$a=7\bigg[-a+\frac{a}{1-r}\bigg].$$ Because $a\neq 0$, we get $$1=7\bigg[-1+\frac{1}{1-r}\bigg].$$ Solving for $r$, we get $r=\frac{1}{8}$. The value $a=16$ follows from the equation $a+ar=18$