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February 1, 2015

Posted by **Watermelon** on Thursday, December 1, 2011 at 8:37am.

(a)the first term and the common ration of the progression

(b)the sum of infinity of the progression

- GP Caluculus -
**bobpursley**, Thursday, December 1, 2011 at 8:51amCalculus, AP?

caluculus, fouth, ration, are not words normally used by Calculus students. Choosing words, and checking the spelling, reflects on you personally. If you desire success, work on these types of details. You can do better than this.

Given that a3 is 16, and a4 is 8, the ratio must be r=.5

a3=a0*r^(3-1)

16=a0*(1/2)^(3-1)or a0=16*4

check: 16*4, 16*2, 16*1, 16*1/2, ...

sum:

Sum= ao/(1-r)=16*4/.5=16*8

check my thinking.

- GP Caluculus -
**Watermelon**, Thursday, December 1, 2011 at 9:09amthis is not arithmetic progression = =

- GP Caluculus -
**Reiny**, Thursday, December 1, 2011 at 9:30amthird term = ar^2 and ar^2 = 16

fourth term = ar^3

ar^2 + ar^3 = 8

16 + ar^3 = 8

ar^3 = -8

ar^3/(ar^2) = -8/16

r = -1/2

then a = 64

terms are: 64 -32 16 -8 4 ...

check: is term3 = 16 ? , YES

is 16 - 8 = 8 ? , yes

sum_{∞}= a/(1-r) = 64/(3/2) = 128/3

- GP Caluculus -
**bobpursley**, Thursday, December 1, 2011 at 10:20amThanks, R. I mis-read the statement, I some how read into the question the fourth term was 8. Duh on me.

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