"In an arithmetic series, a10=3 and S6=76.5. Find a1, d, and the smallest value of n such that Sn<0

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I'm not sure how to start and find a1 or d

Sep 2nd, 2015

Hi there! Thank you for the opportunity to help you with your question!

In an arithmetic series, we have evenly spaced terms (that is, we difference between consecutive terms is constant).

(For example, 4, 7, 10, 13, 16, 19 ... is an arithmetic sequence.)

More generically, if the first term of the sequence is a1, then the second will be

a2 = a1 + d

a3 = a2 + d = a1 + 2d

a4 = a3 + d = a1 + 3d...

and so on

In this problem, we know that the 10th number is 3.

a10 = 3

But we also know from the discussion above that

a10 = a1 + 9d.

So a1 +9d = 3

Now, we need to use the second piece of information, that the sum of the first six terms (S6) is equal to 76.5. Now, how can we write the sum of the first six terms of this series? We could do it explicitly:

S6 = a1+a2+a3+a4+a5+a6 = a1+(a1+d)+(a1+2d)+(a1+3d)+(a1+4d)+(a1+5d) = 6*a1+15d

But we know S6 = 76.5.

So 6*a1 + 15*d = 76.5

Using the two equations in bold we can get that:

6*(3-9*d) +15d = 76.5 

d= -1.5

a1 = 16.5

Let's just see if it makes sense:

16.5 + 15 + 13.5 + 12 +10.5 +9 = 76.5

and if we keep going a7 = 7.5, a8 = 6, a9 = 4.5 a10 =3, as we wanted!

Please let me know if you need any clarification. Always glad to help!
Sep 2nd, 2015

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