Question 6:
There are two geometric progressions, A and B . The first term of A is 8 and the first term of B is 4 . Both progressions have the same sum to infinity. The sum of the 3rd terms of A and B is 9/4. The tenth term of A is negative.
A student states that the sum of the first 4 terms of A is larger than the sum of the first 4 terms of B.
Is the statement correct? Give your justification.
[8 marks]
Answer:
$$ \begin{aligned} S_{\infty \mathrm{A}} & =S_{\infty \mathrm{B}} \\ \frac{8}{1-r_{\mathrm{A}}} & =\frac{4}{1-r_{\mathrm{B}}} \\ 8-8 r_{\mathrm{B}} & =4-4 r_{\mathrm{A}} \\ 4 r_{\mathrm{A}}-8 r_{\mathrm{B}} & =4-8 \\ 4 r_{\mathrm{A}}-8 r_{\mathrm{B}} & =-4 \\ r_{\mathrm{A}}-2 r_{\mathrm{B}} & =-1 \\ r_{\mathrm{A}} & =2 r_{\mathrm{B}}-1 \ldots(1) \end{aligned} $$
$$ \begin{aligned} T_{3 \mathrm{~A}}+T_{3 \mathrm{~B}} & =\frac{9}{4} \\ 8\left(r_{\mathrm{A}}\right)^{3-1}+4\left(r_{\mathrm{B}}\right)^{3-1} & =\frac{9}{4} \\ 8 r_{\mathrm{A}}^2+4 r_{\mathrm{B}}^2 & =\frac{9}{4} \\ 8\left(2 r_{\mathrm{B}}-1\right)^2+4 r_{\mathrm{B}}^2 & =\frac{9}{4} \\ 8\left(4 r_{\mathrm{B}}^2-4 r_{\mathrm{B}}+1\right)+4 r_{\mathrm{B}}^2-\frac{9}{4} & =0 \\ 32 r_{\mathrm{B}}^2-32 r_{\mathrm{B}}+8+4 r_{\mathrm{B}}^2-\frac{9}{4} & =0 \\ 36 r_{\mathrm{B}}^2-32 r_{\mathrm{B}}+\frac{23}{4} & =0 \\ 36 r_{\mathrm{B}}^2-32 r_{\mathrm{B}}+\frac{23}{4} & =0 \end{aligned} $$
$$ \begin{aligned} & 144 r_{\mathrm{B}}^2-128 r_{\mathrm{B}}+23=0 \\ &\left(36 r_{\mathrm{B}}-23\right)\left(4 r_{\mathrm{B}}-1\right)=0 \\ & 36 r_{\mathrm{B}}-23=0, \quad 4 r_{\mathrm{B}}-1=0 \\ & r_{\mathrm{B}}=\frac{23}{36}, \quad r_{\mathrm{B}}=\frac{1}{4} \\ & T_{10 \mathrm{~A}}<0 \\ & 8\left(r_{\mathrm{A}}\right)^{10-1}<0 \\ & 8\left(r_{\mathrm{A}}\right)^9<0 \\ & r_{\mathrm{A}}<0 \ldots(2) \end{aligned} $$
$$ \begin{aligned} &\text { From (1) & (2), }\\ &\begin{aligned} & r_{\mathrm{A}}=2\left(\frac{23}{36}\right)-1 \quad, \quad r_{\mathrm{A}}=2\left(\frac{1}{4}\right)-1 \\ & r_{\mathrm{A}}=\frac{5}{18}<0 \text { (ignore) } \quad, \quad r_{\mathrm{A}}=-\frac{1}{2}<0 \end{aligned}\\ &\begin{aligned} S_{4 \mathrm{~A}} & =\frac{8\left[1-\left(-\frac{1}{2}\right)^4\right]}{1-\left(-\frac{1}{2}\right)} \\ & =5<S_{4 \mathrm{~B}} \\ S_{4 \mathrm{~B}} & =\frac{4\left[1-\left(\frac{1}{4}\right)^4\right]}{1-\left(\frac{1}{4}\right)}=5.3125 \end{aligned} \end{aligned} $$
∴ The statement is false.
There are two geometric progressions, A and B . The first term of A is 8 and the first term of B is 4 . Both progressions have the same sum to infinity. The sum of the 3rd terms of A and B is 9/4. The tenth term of A is negative.
A student states that the sum of the first 4 terms of A is larger than the sum of the first 4 terms of B.
Is the statement correct? Give your justification.
[8 marks]
Answer:
$$ \begin{aligned} S_{\infty \mathrm{A}} & =S_{\infty \mathrm{B}} \\ \frac{8}{1-r_{\mathrm{A}}} & =\frac{4}{1-r_{\mathrm{B}}} \\ 8-8 r_{\mathrm{B}} & =4-4 r_{\mathrm{A}} \\ 4 r_{\mathrm{A}}-8 r_{\mathrm{B}} & =4-8 \\ 4 r_{\mathrm{A}}-8 r_{\mathrm{B}} & =-4 \\ r_{\mathrm{A}}-2 r_{\mathrm{B}} & =-1 \\ r_{\mathrm{A}} & =2 r_{\mathrm{B}}-1 \ldots(1) \end{aligned} $$
$$ \begin{aligned} T_{3 \mathrm{~A}}+T_{3 \mathrm{~B}} & =\frac{9}{4} \\ 8\left(r_{\mathrm{A}}\right)^{3-1}+4\left(r_{\mathrm{B}}\right)^{3-1} & =\frac{9}{4} \\ 8 r_{\mathrm{A}}^2+4 r_{\mathrm{B}}^2 & =\frac{9}{4} \\ 8\left(2 r_{\mathrm{B}}-1\right)^2+4 r_{\mathrm{B}}^2 & =\frac{9}{4} \\ 8\left(4 r_{\mathrm{B}}^2-4 r_{\mathrm{B}}+1\right)+4 r_{\mathrm{B}}^2-\frac{9}{4} & =0 \\ 32 r_{\mathrm{B}}^2-32 r_{\mathrm{B}}+8+4 r_{\mathrm{B}}^2-\frac{9}{4} & =0 \\ 36 r_{\mathrm{B}}^2-32 r_{\mathrm{B}}+\frac{23}{4} & =0 \\ 36 r_{\mathrm{B}}^2-32 r_{\mathrm{B}}+\frac{23}{4} & =0 \end{aligned} $$
$$ \begin{aligned} & 144 r_{\mathrm{B}}^2-128 r_{\mathrm{B}}+23=0 \\ &\left(36 r_{\mathrm{B}}-23\right)\left(4 r_{\mathrm{B}}-1\right)=0 \\ & 36 r_{\mathrm{B}}-23=0, \quad 4 r_{\mathrm{B}}-1=0 \\ & r_{\mathrm{B}}=\frac{23}{36}, \quad r_{\mathrm{B}}=\frac{1}{4} \\ & T_{10 \mathrm{~A}}<0 \\ & 8\left(r_{\mathrm{A}}\right)^{10-1}<0 \\ & 8\left(r_{\mathrm{A}}\right)^9<0 \\ & r_{\mathrm{A}}<0 \ldots(2) \end{aligned} $$
$$ \begin{aligned} &\text { From (1) & (2), }\\ &\begin{aligned} & r_{\mathrm{A}}=2\left(\frac{23}{36}\right)-1 \quad, \quad r_{\mathrm{A}}=2\left(\frac{1}{4}\right)-1 \\ & r_{\mathrm{A}}=\frac{5}{18}<0 \text { (ignore) } \quad, \quad r_{\mathrm{A}}=-\frac{1}{2}<0 \end{aligned}\\ &\begin{aligned} S_{4 \mathrm{~A}} & =\frac{8\left[1-\left(-\frac{1}{2}\right)^4\right]}{1-\left(-\frac{1}{2}\right)} \\ & =5<S_{4 \mathrm{~B}} \\ S_{4 \mathrm{~B}} & =\frac{4\left[1-\left(\frac{1}{4}\right)^4\right]}{1-\left(\frac{1}{4}\right)}=5.3125 \end{aligned} \end{aligned} $$
∴ The statement is false.