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Calculate $n$ if $\sum_{k=1}^{n} (4 - 3k) = -125$
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T1= 4-3(1)=1

T2=4-3(2)= -2

T3=4-3(3)= -5

T4= 4-3(4)= -8

There is common difference , -2-1= -5-(-2)= -3

sum of arithmetic = n(2a+(n-1)d)/2

-125= 20(2(1)+(n-1)(-3))/2

-12.5 = 2-3n+3

3n= 12.5+2+3

3n= 17.5

n= 8.83
by Diamond (45,616 points)

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