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How do I do Core 1 Ex 6H Q7?

Find the least value of n for which n
Z (4r-3) > 2000
r=1


Z is that sideways M thing
Original post by Maths_Struggles
Find the least value of n for which n
Z (4r-3) > 2000
r=1


Z is that sideways M thing


I believe you mean Sigma as the 'M thing' (Summation)

What have you done so far?
I don't know where to start. I havent seen > in a question yet either.
The past Qs I have done are like find the sequence, so I know what n is etc
Original post by Maths_Struggles
I don't know where to start. I havent seen > in a question yet either.
The past Qs I have done are like find the sequence, so I know what n is etc


Well basically you're looking for the amount of terms where the sum of the terms is greater than 2000.

If it helps work out the first few terms so U1 = 1 , U2 = 5, U3 = 9 This also gives you a and d.

Obviously the sum of the first 3 terms is 15 so you are miles off being greater than 2000. Think your formulas for working out the sum as it would be too time consuming obviously to work it out by hand.
Thanks :smile:
1/2n(2a+(n-1)d ? That formula?
Original post by Maths_Struggles
Thanks :smile:
1/2n(2a+(n-1)d ? That formula?

Yes, that would work.
Thanks for the help :biggrin: how do I find n though?
Original post by Maths_Struggles
Thanks for the help :biggrin: how do I find n though?

Well, you have a and d, so sub them into n2(2a+(n1)d)\frac{n}{2}(2a+(n-1)d), and you know that this must be greater than 2000. Could you now rearrange to find n?
Yeh got it thanks everyone :biggrin:

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