You may use the following formula, such that:
`a_n = a_1*r^(n-1)`
Replacing 8 for n,`1/2 ` for` a_1` yields:
`a_8 = 1/2*(r)^7`
You may find the ratio r, such that:
`r = (a_2)/(a_1) => r = (-1/8)/(1/2) => r = -1/4`
`a_8 = 1/2*(-1/4)^7 => a_8 = -1/(2^15)`
Hence, the indicated term `a_8 ` is `a_8 = -1/(2^15)`
You may use the following formula, such that:
`a_n = a_1*r^(n-1)`
Replacing 8 for n,`1/2 ` for` a_1` yields:
`a_8 = 1/2*(r)^7`
You may find the ratio r, such that:
`r = (a_2)/(a_1) => r = (-1/8)/(1/2) => r = -1/4`
`a_8 = 1/2*(-1/4)^7 => a_8 = -1/(2^15)`
Hence, the indicated term `a_8 ` is `a_8 = -1/(2^15)`
No comments:
Post a Comment