Answer:
a). r = [tex]\frac{6}{7}[/tex]
b). At least 5 terms should be added.
Step-by-step explanation:
Formula representing sum of infinite geometric sequence is,
[tex]S_{\inf}=\frac{a}{1-r}[/tex]
Where a = first term of the sequence
r = common ratio
a). If the sum is seven times the value of its first term.
[tex]7a=\frac{a}{1-r}[/tex]
[tex]7=\frac{1}{1-r}[/tex]
7(1 - r) = 1
7 - 7r = 1
7r = 7 - 1
7r = 6
r = [tex]\frac{6}{7}[/tex]
b). Since sum of n terms of the geometric sequence is given by,
[tex]S_{n}=\frac{a(1-r^{n})}{1-r}[/tex]
If the sum of n terms of this sequence is more than half the value of the infinite sum.
[tex]\frac{a[1-(\frac{6}{7})^{n}]}{1-\frac{6}{7}}[/tex] > [tex]\frac{7a}{2}[/tex]
[tex]\frac{1-(\frac{6}{7})^{n}}{1-\frac{6}{7}}> \frac{7}{2}[/tex]
[tex]\frac{1-(\frac{6}{7})^{n}}{\frac{1}{7}}> \frac{7}{2}[/tex]
[tex]1-(\frac{6}{7})^{n}> \frac{7}{2}\times \frac{1}{7}[/tex]
[tex]1-(\frac{6}{7})^{n}> \frac{1}{2}[/tex]
[tex]-(\frac{6}{7})^{n}> -\frac{1}{2}[/tex]
[tex](\frac{6}{7})^{n}< \frac{1}{2}[/tex]
[tex](0.85714)^{n}< (0.5)[/tex]
n[log(0.85714)] < log(0.5)
-n(0.06695) < -0.30102
n > [tex]\frac{0.30102}{0.06695}[/tex]
n > 4.496
n > 4.5
Therefore, at least 5 terms of the sequence should be added.
