what is the sumof the series? 20 n-1 (5-n) A.-220 B.-110 C.-95 D.110
what is the 10th term of the geometric sequence 400,200,100...? A.0.09765625 B.0.390625 C.0.78125 D.1.5625
Hold on a sec, one at a time :) Let's start with the first one.
Just to double check, the series is this: \[20 n-1 (5-n)\]Yes?
the pressure,P, of a gas varies directly with its temperature,T, and inversely with its volume,V, according to the equation P=nRT/V, where n is the number of molar units and R is the universal gas constant. Onemolar unit of a gas has a pressure of the same number of molar units of the gas at a temperature of 400 Kelin and a volume of 2.5 liters? A.1,328 joules B.1,900 joules C.2,375 joules D.2,600 joules
yes @tom982
The sum up to what number? It's infinity otherwise
is it -110 @tom982
Is there not more to this question? You can't just find the sum of an arithmetic series. I'm expecting it will say something like "find the sum of this series up to n=20".
@tom982 I think (this is just a guess really) that `20 n-1 (5-n)` is really saying "sum (5-n) from n = 1 to n = 20" so it would be written as \[\Large \sum_{n=1}^{20}(5-n)\]
Ah that makes a lot more sense. I did think it was written a bit oddly. @Saylilbaby can you just confirm this is the case before we continue? Also, can you tell me the formula for the sum of an arithmetic series?
\[\sum_{n-1}^{20} (5-n)\]
yes @jim_thompson5910 and @tom982
Great. Did you manage to find the formula for the sum of an arithmetic series?
{a,a+d,a+2d,a+3d,...} @tom982
I've got to go now so I'll leave you with enough information to solve this, then someone else will hopefully be able to verify your answer. The sum is defined as:\[\dfrac{n}{2}(2a+(n-1)d)\]Where: n is the number of terms, so n=20 a is the first term, so a=(5-1)=4 d is the common difference, so d=-1 Put those values into the equation and you'll get your answer.
You could do it like you said, by adding up each term, but it's tedious and can take a long time with lots of terms.
i got 260 but that isnt an answer @tom982
\[\dfrac{n}{2}(2a+(n-1)d)=\dfrac{20}{2}(2*4+(20-1)*(-1)=10(8-19)=10*-11-220\]
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