Question

The Second, Fourth and Seventh terms of an arithmetic series are the first 3 consecutive

terms of a geometric series. Find
(a) The common ratio. (6 marks)
(b) The sum of the first six terms of the geometric series if the common difference of the
arithmetic series is 2. (4 marks)

Answer 1

Hi,

Explanation:

Let's assume a the first term of the arithmetic serie

a+2 the second term

a+6 the fourth term

a+12 the seventh term.

Let's assume r the ratio.

`a+6=(a+2)*r\\a+12=(a+6)*r\\\\(a+12)/(a+6)=(a+6)/(a+2) \Longrightarrow\ (a+12)*(a+2)=(a+6)^2\\\\\Longrightarrow\ a^2+12a+2a+24=a^2+12a+36\\\\\Longrightarrow\ 2a=12\\\\\Longrightarrow\ a=6\\\\a)\\r=(a+6)/(a+2) =(12)/(8) =(3)/(2) \\\\`

`b)\\b=a+2=8\\\\b+b*r+b*r^2+b*r^3+...+b*r^5\\\\=b*(r^6-1)/(r-1) \\\\=8*(((3)/(2))^6 -1))/((1)/(2))\\\\=(665)/(4) \\\\=166.25`