Asked by Anonymous
If the first and the last terms of an arithmetic series are 10 and 62, show that the sum of the series varies directly as the number of terms
Answers
Answered by
Reiny
a=10
a+(n-1)d = 62
10 + nd - d = 62
nd-d=52
n=(52+d)/d or d = 52/(n-1)
sum(n) = (n/2)(2a + (n-1)d)
= (n/2)(20 + (n-1)(52)/(n-1))
= (n/2)(20 + 52)
= (n/2)(72) = 36n
sum(n) = 36n
so sum(n) varies directly as the number of terms
a+(n-1)d = 62
10 + nd - d = 62
nd-d=52
n=(52+d)/d or d = 52/(n-1)
sum(n) = (n/2)(2a + (n-1)d)
= (n/2)(20 + (n-1)(52)/(n-1))
= (n/2)(20 + 52)
= (n/2)(72) = 36n
sum(n) = 36n
so sum(n) varies directly as the number of terms
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