Asked by Lana
Find the sum of the 50 greatest negative integers.
I'm supposed to find a, d, and t(at)n, but I'm not sure which is which. I think it makes sense for n to equal 50, but apart from that, I don't know what else to do. I would really appreciate some sort of explanation on how to solve this. Thank you. Oh, and I have a test on this tomorrow, so I really need the help. :)
I'm supposed to find a, d, and t(at)n, but I'm not sure which is which. I think it makes sense for n to equal 50, but apart from that, I don't know what else to do. I would really appreciate some sort of explanation on how to solve this. Thank you. Oh, and I have a test on this tomorrow, so I really need the help. :)
Answers
Answered by
Reiny
you are simply asked to find
(-50) + (-49) + ... + (-2) + (-1)
there are 50 of these, so n=50
the first term is -50, so a=-50
the common difference is 1, so d=1
S<sub>50</sub> = 50/2(-50 + (-1) = -1275
I was using S(n) = n/2(first + last)
You could have used
S(n) = n/2[2a + (n-1)d]
= 25[-100 + 49]
= -1275
notice if you had added up the first 50 positive integers you would get +1275
(-50) + (-49) + ... + (-2) + (-1)
there are 50 of these, so n=50
the first term is -50, so a=-50
the common difference is 1, so d=1
S<sub>50</sub> = 50/2(-50 + (-1) = -1275
I was using S(n) = n/2(first + last)
You could have used
S(n) = n/2[2a + (n-1)d]
= 25[-100 + 49]
= -1275
notice if you had added up the first 50 positive integers you would get +1275
Answered by
Lana
oh! a = -50?!
for some reason I thought it was 1, but it makes much more sense now. Okay, yeah, I see what you mean. :) Thank you so very much for helping me! :D
for some reason I thought it was 1, but it makes much more sense now. Okay, yeah, I see what you mean. :) Thank you so very much for helping me! :D
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