Question

Find S sub 12 for the geometric series:
1.5+ (-3) + 6 +....

Answer 1

S12 for geometric series: 1.5+ (-3) + 6 +.... would be: -2047.5

Explanation:

Given the sequence to find the sum up-to 12 terms

`1.5+ (-3) + 6 +....`

A geometric sequence has a constant ratio 'r' and is defined by

`a_n=a_1\cdot r^(n-1)`

`\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\q\:r=(a_(n+1))/(a_n)`

`(\left(-3\right))/(1.5)=-2,\:\quad (\left(6\right))/(\left(-3\right))=-2`

`\mathrm{The\:ratio\:of\:all\:the\:adjacent\:terms\:is\:the\:same\:and\:equal\:to}`

`r=-2`

`\mathrm{The\:first\:element\:of\:the\:sequence\:is}`

`a_1=1.5`

as

`a_n=a_1\cdot r^(n-1)`

`\mathrm{Therefore,\:the\:}n\mathrm{th\:term\:is\:computed\:by}\:`

`a_n=1.5\left(-2\right)^(n-1)`

`\mathrm{Geometric\:sequence\:sum\:formula:}`

`a_1(1-r^n)/(1-r)`

`\mathrm{Plug\:in\:the\:values:}`

`n=12,\:\spacea_1=1.5,\:\spacer=-2`

`=1.5\cdot (1-\left(-2\right)^(12))/(1-\left(-2\right))`

`=1.5\cdot (1-\left(-2\right)^(12))/(1+2)`

`\mathrm{Multiply\:fractions}:\quad \:a\cdot (b)/(c)=(a\:\cdot \:b)/(c)`

`=(\left(1-\left(-2\right)^(12)\right)\cdot \:1.5)/(1+2)`

`=(-6142.5)/(1+2)` ∵
`\left(1-\left(-2\right)^(12)\right)\cdot \:1.5=-6142.5`

`=(-6142.5)/(3)`

`\mathrm{Apply\:the\:fraction\:rule}:\quad (-a)/(b)=-(a)/(b)`

`=-(6142.5)/(3)`

`=-2047.5`

Thus, S12 for geometric series: 1.5+ (-3) + 6 +.... would be: -2047.5