Question

The sum of an infinite geometric sequence is seven times the value of its first term.

Answer 1

Answer:

a). r = $\frac{6}{7}$

b). At least 5 terms should be added.

Step-by-step explanation:

Formula representing sum of infinite geometric sequence is,

$S_{\inf}=\frac{a}{1-r}$

Where a = first term of the sequence

r = common ratio

a). If the sum is seven times the value of its first term.

    $7a=\frac{a}{1-r}$

    $7=\frac{1}{1-r}$

    7(1 - r) = 1

    7 - 7r = 1

    7r = 7 - 1

    7r = 6

    r = $\frac{6}{7}$

b). Since sum of n terms of the geometric sequence is given by,

    $S_{n}=\frac{a(1-r^{n})}{1-r}$

If the sum of n terms of this sequence is more than half the value of the infinite sum.

$\frac{a[1-(\frac{6}{7})^{n}]}{1-\frac{6}{7}}$ >  $\frac{7a}{2}$

$\frac{1-(\frac{6}{7})^{n}}{1-\frac{6}{7}}> \frac{7}{2}$

$\frac{1-(\frac{6}{7})^{n}}{\frac{1}{7}}> \frac{7}{2}$

$1-(\frac{6}{7})^{n}> \frac{7}{2}\times \frac{1}{7}$

$1-(\frac{6}{7})^{n}> \frac{1}{2}$

$-(\frac{6}{7})^{n}> -\frac{1}{2}$

$(\frac{6}{7})^{n}< \frac{1}{2}$

$(0.85714)^{n}< (0.5)$

n[log(0.85714)] < log(0.5)

-n(0.06695) < -0.30102

n > $\frac{0.30102}{0.06695}$

n > 4.496

n > 4.5

Therefore, at least 5 terms of the sequence should be added.