+0

# ​Help me with these geomteric sequence questions

0
157
1 help

Feb 25, 2021

#1
+1

3. (a) We know that $a_3 = 45$ and $a_5 = 101.25.$ If we express $a_3 = a_1 r^2$ and $a_5 = a_1 r^4$, we can find the common ratio if we divide $a_5/a_3$. Doing this, we get $$r^2 = \frac{a_5}{a_3},$$ $$r^2 = \frac{101.25}{45} = 2.25.$$ Since we know that all of the terms are positive, the common ratio $r=1.5,$ and we find the first term by substituting the common ratio into the equation for $a_3$. We get $$45 = a_1 (1.5)^2,$$ $$a_1 =20.$$ Therefore, our answer is $\boxed{a_1 = 20, r=1.5}.$

(b) We find the largest term greater than $2000$ by expressing $a_x = a_1 r^{x-1}.$ Given that $r=1.5$ (from part(a)), we find that $x-1 = 12$. Therefore, the thirteenth term exceeds $2000$, and $\boxed{a_{13} \approx 2594.93}.$

5. (a) We know three successive terms of a geometric sequence are $x-6,x,2x+9$. We can express these terms like we did before, and we get $$x-6 = a r,$$ $$x = ar^2,$$ $$2x+9 = ar^3.$$ Therefore, we can solve two equations for $r$, and we get $$r = \frac{x}{x-6} = \frac{2x+9}{x}.$$ Solving for $x$, we get $(x-9)(x+6)=0$. Since all terms in the sequence must be positive, our answer is $\boxed{x=9}$.

(b) Since $x$ is the $5$th term, $a_5 = x = a_1 r^4.$ We can solve for $r$ from the equations before, where $r = \frac{x}{x-6}.$ We know that $x=9$, so $r=\frac{9}{3} = 3.$ Therefore, we know $9 = a_1 (3)^4.$ Therefore, we get $\boxed{a_1 = \frac{1}{9}}.$

6. (a) Since Nurhaniza was her salary increase $12$ percent every year, her monthly salary after 6 years would just be $$\text{final salary = starting salary }\cdot \text{(annual increase %)}^6.$$ Plugging in our values for the starting salary and the annual increase %, we get $$\text{final salary} = RM1500 \cdot (112 \text{%})^6,$$ $$\text{final salary} = RM2960.73.$$

(b) Part (b) asks us for the number of years that she must stay at the company until her monthly salary exceeds $RM3500.$ Therefore, we are asked to find $n$ in the inequality $RM1500 \cdot (112 \text{%})^n > RM3500.$ Solving this inequality, we find that $n=8.$ Therefore, it takes $\boxed{8}$ years with the company for Nurhaniza's monthly salary to exceed $RM3500.$