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anonymous
 one year ago
The 1st term of an arithmetic progression is a and the common difference is d, where d does not equal 0.
(i) Write down expressions, in terms of a and d, for the 5th term and the 15th term.
The 1st term, the 5th term and the 15th term of the arithmetic progression are the first three terms of a geometric progression.
(ii) Show that 3a=8d
I need help with (ii) please! I have the equations for (i) they are a + 4d ; a +14d
anonymous
 one year ago
The 1st term of an arithmetic progression is a and the common difference is d, where d does not equal 0. (i) Write down expressions, in terms of a and d, for the 5th term and the 15th term. The 1st term, the 5th term and the 15th term of the arithmetic progression are the first three terms of a geometric progression. (ii) Show that 3a=8d I need help with (ii) please! I have the equations for (i) they are a + 4d ; a +14d

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DanJS
 one year ago
Best ResponseYou've already chosen the best response.2have you done anything yet?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I actually don't even know where to start so no.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2first, you have to figure what an arithmetic progression is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well I know what that is already

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2Arithmetic sequence of numbers starts at a, and each consecutive term is 'd' away from the last term

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2a general form for the nth term is, \[a _{n} = a _{1} + (n  1)*d\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, i have the answers for the first question already figured out, its the second one i would like help with please

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2the nth term = the first term + (n1) d

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2do you know what a geometric series is?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2instead of an linear change from term to term, geometric series multiplies each by a common ratio r, every term / the one before it = r , and it is constant

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2The three terms from the last series to use are a ; a + 4d ; a + 14d

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2the common ratio between terms is r = (a + 4d) / a and r = (a + 14d)/(a+4d)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2general geometric series to get the nth term Nth Term = (FIrst Term) * r^n

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2really all you have to do is know that the ratio of consecutive terms is always the same value r

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2set those two ratios equal, and solve for 3a to show it equals 8d

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ a+4d }{ a } = \frac{ a+14d }{ a+4d }\] = r

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2solve for 3a = ... should come to that 8d value

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok when you say solve for 3a how would I do that exactly (sorry)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2just start moving things around and expanding and isolating a to one side and d to the other side

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2cross multiply first, for example

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2(a+4d)*(a+4d) = a*(a+14d)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2expand all those out from disributing

DanJS
 one year ago
Best ResponseYou've already chosen the best response.2should be left with what they are looking for

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, will try that quickly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hooray I got the right answer! Thank you very much for your help!
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