Feellar 0.

asked • 11/01/17

can someone help me?

the first, fourth and thirteenth terms of an arithmetic series are consecutive terms in a (non-constant) geometric series. The sixth term in the arithmetic series is 78. find the first term and the common difference of the arithmetic series?

1 Expert Answer

By:

Mark M. answered • 11/01/17

Tutor
4.9 (954)

Retired math prof. Very extensive Precalculus tutoring experience.

About this tutor ›
About this tutor ›

answer withdrawn--misread the question.

Upvote • 0 Downvote
More
Report

Feellar 0.

Hello. 
Sorry, i think you made mistake on that. you can't say that r = 78/a because 78 is only for AP which is a+5d?
 
am I correct?
Report

11/01/17

Mark M.

tutor
It's too difficult to edit previously posted answers, so I am entering the answer as a comment:
 
a1 = a    a4 = a+3d = ar     a13 = a+12d = ar2
 
Since a6 = a + 5d = 78, d = (78-a)/5
 
So, a + (3/5)(78-a) = ar.  Therefore, r = 1 + (3/(5a))(78-a)
 
So, a + (12/5)(78-a) = a[1 + (3/(5a))(78-a)]2
 
     a + (12/5)(78-a) = (1/25a)[234+2a)2
 
     25a2 + 60a(78-a) = (234+2a)2
 
Simplify to obtain 39a2 -3744a + 54756 = 0
 
The quadratic formula gives a = 18 or 78.
 
If a = 78, then r = 1, which makes the geometric sequence constant.  
 
So, a = 18 and d = (78-18)/5 = 12.  
 
Report

11/01/17

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.