What is the sum of the numbers in the series? 15 + 11 + 7 + . . . + (β129)A. -2,124B. -2,109C. -2,052D. -1,995
Question
Answer:
Using Gauss's methodTotal number of terms = [15-(-129)]/4+1=36+1=37
Add
S=15+11+7+....-125-129
S=-129-125-...+7+11+15
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2S=-114-114-114...(37 times)
=>
sum=S=(1/2)*(-114)*37=-2109
Using AP, T(n)=15+11+7+....-129
T(n)=19-4nΒ => T(1)=15, T(37)=-129
S(n)=(1/2)(37)(T(1)+T(37)=(1/2)37(15-129)=2109
solved
general
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