Sequence and Series (1 Viewer)

fullonoob · Jan 5, 2010

<DIR>Find the common ratio of the GP in which the first, third and fourth terms form an arithmetic sequence. [HINT: r^3-2r^2 +1 = (r-1)(r^2-r-1)]
WTFUXX :mad1:
</DIR>

untouchablecuz · Jan 5, 2010

fullonoob · Jan 5, 2010

untouchablecuz said:
$\\\text{The general term of a geometric progression is: }T_n=ar^{n-1}\\T_1=ar^{1-1}=a\\T_3=ar^{3-1}=ar^2\\T_4=ar^{4-1}=ar^3\\\text{If the terms in question are in a arithmetic series, then, }T_3-T_1=T_4-T_3\\\text {i.e. }ar^2-a=ar^3-ar^2\Rightarrow r^3-2r^2+1=0\\\text{Given that } r^3-2r^2+1 = (r-1)(r^2-r-1)\\\therefore r=1\text{ or } \frac{1\pm \sqrt{5}}{2}$

I see, but i dont get how you can assume a = 1? Explain please

EDIT LOLOLOL DW THE As cancel out my bad

Eddykungfu · Jan 5, 2010

so glad that q wasn't in the 09 MX1 paper

i would have fml even more.

Sequence and Series (1 Viewer)

fullonoob

fail engrish? unpossible!

untouchablecuz

Active Member

fullonoob

fail engrish? unpossible!

Eddykungfu

Member

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