Thread: Arithmetic Series problem

In an arithmetic sequence, the 6th term is half the fourth term and the third term is 15.
i) Find the first term and the common difference
ii) How many terms are needed to give a sum that is less than 65?

I've done part (i) and found a=21 and d=-3, but I'm having trouble with the second part. Here's what I've got so far:

S_n = n/2{2a+(n-1)d}
S_n = n/2{2(21) + (n-1)(-3)}

S_n = n/2{45-3n}
(n/2){45-3n} < 65
n{45-3n} < 130
45n -3n² < 130

I think I'm on the right track (I think???)...but I don’t think the quadratic inequality isn't giving me whole numbers if I solve for values of n so I’ve gone wrong somewhere. Any help? Thanks

Are you sure you have written the problem correctly? Yes, a= 21 and d= -3 so the first few terms are 21, 18, 15, .... The "sum" of the first one term (if you call that a sum) is 21, the sum of the first two is 39, the sum of the first three is 54. Those are all less than 65! The sum of the first four terms is 21+ 18+ 15+ 12= 66 so that after the fourth term the sum is NO LONGER less than 65. Of course, and beyond are negative so that the sum begins decreasing and will eventually be less than 65 again. Was that what was meant?

As for "I don’t think the quadratic inequality isn't giving me whole numbers if I solve for values of n", of course it does! The equation 45n- 3n²= 130 has no integer roots but 45n- 3n²< 130 does. The roots of 45n- 3n²= 130 are a little smaller than 4 and a little larger than 11. For all integers between 4 and 11, 5, 6, 7, 8, 9, and 10, 45n- 3n²> 130 so that sum is larger than 65. For all n larger than 11, the sum is again less than 65.

So now it is enough to solve this quadratic inequality. you may consider the following equivalent inequality
solving the quadratic equation to get the zeros (11.09... and 3.90...) and outside the zeros the quadratic equation is positive
so your series is less than 65 until the third term and from the 12th term

Ah yes I get it now! Thank you!