>> Prove that no matter what the real numbers 'a' and 'b' are, the sequence with nth term a+nb is always an A.P. What is the sum of the first 20 terms of this sequence?

 

plz show the steps if possible

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The difference between consecutive terms is a+nb-(a+(n-1)b) = a+nb-a-nb+b = b. So there is a fixed number b between each term, which is the definition of an arithmetic progression, each term increasing by the same amount from term to term. The first term in the sequence would be a if the sequence starts at n=0, or a+b if it starts at n=1, when the nth term would be a+20b. So the series goes: a+b, a+2b, a+3b, ... a+20b. The sum of n terms is 20a (because we're adding a 20 times) + (20*21/2)b = 20a+210b. The easiest way to find the sum of an arithmetic series is to note that the first and last terms add up to 2a+21b, the second term and penultimate term also add up to this, and so on. We only have to go halfway so we only add 10 complementary terms together, so that's 10(2a+21b), which is the answer above. The formula is n(n+1)/2.

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