The number of citizens is rising in the town by the same/equal percentage each year in contrast to the previous year. In 6 years time, the number of citizens rose from 1 635000 to 2 010000. What is the percentage of growth each year equal to?
My solution:
$$a_6=a_1*q^{n-1}; n=6$$ $$2 010000=1635000*q^5 => q=\sqrt[\leftroot{-2}\uproot{5}5]{\frac{2010000}{1635000}}$$ and then I get: $$q=1.042$$
Can anyone help on this one..
The result should be 3.5%
Since 1635000 is the non increased value therefore it is the $a_0$ or the $a_1$ but then we add $a_7$ as the final value.
Therefore:
$$a_7=a_1*q^6 => q=1.035 => 0.035$$ That's 0.035 increase or 3.5% each year.
In other words:
$$a_1=0x\space increase=1,635000 $$ $$a_2=1x\space increase$$ $$a_3=2x\space increase$$ $$a_4=3x\space increase$$ $$a_5=4x\space increase$$ $$a_6=5x\space increase$$ $$a_7=6x\space increase=2,010000$$
Some other way to solve this problem is through recursion: $$T(n)=T(n-1)(1+a)$$ $$T(0)=b$$ where $b$ is the basic number of citizens and $a$ is the fraction of increase. Now the solution of this recursion is $$b \times (1+a)^n$$ For $$T(6)=b \times (1+a)^6=2010000\implies (1+a)^6=\frac{2010000}{1635000}$$ Now solve for a.
$$B=A\;(1+g)^6\\ g=\left(\frac BA\right)^\frac16-1 $$
