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Sagot :
Answer:
a) Common ratio = 2/7
b) First term = 135/7
Explanations:
The formula for finding the sum of a geometric progression is expressed as:
[tex]S_n=\frac{a(r^n-1)}{r-1}[/tex]Since the sum of the first two terms is 15, then
S2 = 15
n = 2
Substitute into the formula:
[tex]\begin{gathered} S_2=\frac{a\mleft(r^2-1\mright)}{r^{}-1} \\ 15=\frac{a(r+1)\cancel{r-1}}{\cancel{r-1}} \\ 15=a(r+1) \end{gathered}[/tex]Also, the sum to infinity of a geometric sequence is expressed as:
[tex]\begin{gathered} S_{\infty}=\frac{a}{1-r} \\ _{} \end{gathered}[/tex]Substitute the given values into the formula:
[tex]27=\frac{a}{1-r}[/tex]Solve both expressions simultaneously
[tex]\begin{gathered} 15=a(r+1) \\ 27=\frac{a}{1-r} \end{gathered}[/tex]
Divide both expressions to have:
[tex]\frac{15}{27}=\frac{1-r}{r+1}[/tex]Cross multiply and solve for the common ratio "r"
[tex]\begin{gathered} 15(r+1)=27(1-r) \\ 15r+15=27-27r \\ 15r+27r=27-15 \\ 42r=12 \\ r=\frac{12}{42} \\ r=\frac{2}{7} \end{gathered}[/tex]Hence the value of the common ratio is 2/7
b) Get the first term of the sequence;
Using the formula:
[tex]\begin{gathered} 27=\frac{a}{1-r} \\ 27=\frac{a}{1-\frac{2}{7}} \\ 27=\frac{a}{(\frac{5}{7})} \\ a=27\times\frac{5}{7} \\ a=\frac{135}{7} \\ \end{gathered}[/tex]Hence the first term of the sequence is 135/7
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