Question 878423
{{{A[n]=A0+(n-1)R}}}
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{{{A[10]=A0+(10-1)R}}}
{{{A0+9R=20}}}
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{{{A[5]=A0+(5-1)R}}}
{{{A0+4R=0}}}
Subtract these two terms,
{{{A0+9R-A0-4R=20-0}}}
{{{5R=20}}}
{{{R=4}}}
Then,
{{{A0+4(4)=0}}}
{{{A0=-16}}}
So,
{{{A[n]=-16+4(n-1)}}}
{{{A[n]=-16+4n-4}}}
{{{A[n]=4n-20}}}
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{{{sum(A[n],1,N)=sum((4n-20),n=1,N)}}}
{{{sum(A[n],1,N)=4*sum(n,n=1,N)+N(-20)}}}
{{{sum(A[n],1,N)=4*(N(N+1)/2)+N(-20)}}}
{{{sum(A[n],1,N)=2N(N+1)-20N}}}
{{{sum(A[n],1,N)=2N^2+2N-20N}}}
{{{S[N]=sum(A[n],1,N)=2N^2-18N}}}
{{{S[18]=2(18)^2-18*18}}}
{{{S[18]=324}}}
{{{S[7]=2(7)^2-18(7)}}}
{{{S[7]=98-126}}}
{{{S[7]=-28}}}
So the the sum of the terms from 8th to 18th is,
{{{S=S[18]-S[7]=324-(-28)=352}}}