Question 1162188

an arithmetic progression:

{{{a[1]}}}, {{{a[1]+d}}}, {{{a[1]+2d}}},.... where {{{a[1]}}} is the first term and {{{d}}} is Common difference


nth term formula:


{{{a[n]=a[1]+d(n-1)}}}


Denote this partial sum by {{{S[n]}}} . Then

{{{S[n]=n(a[1]+a[n])/2}}} ,where {{{n}}} is the number of terms, {{{a[1]}}} is the first term and {{{a[n]}}} is the last term


given: the sum of the first ten terms is {{{50 }}}

-> {{{n=10}}}

{{{50=10(a[1]+a[10])/2}}}

{{{50=5(a[1]+a[10])}}}

{{{50=5(a[1]+a[10])}}}

{{{10=a[1] + a[10]}}}.......solve for {{{a[1]}}}

{{{a[1]=10-a[10] }}}....{{{a[10] =a[1]+9d}}}

{{{a[1]=10-(a[1]+9d)}}}

{{{a[1]=10-a[1]-9d}}}

{{{a[1]+a[1]=10-9d}}}

{{{2a[1]=10-9d}}}

{{{a[1]=(10-9d)/2}}}.....eq.1



the 5th term is three times the second term:

{{{a[5]=3*a[2]}}}->{{{a[2]=a[1]+d}}} and {{{a[5]=a[1]+4d}}}


{{{a[1]+4d=3(a[1]+d)}}}

{{{a[1]+4d=3a[1]+3d}}}

{{{4d-3d=3a[1]-a[1]}}}

{{{d=2a[1]}}}......... ..eq.2


substitute in eq1

{{{a[1]=(10-9(2a[1]))/2}}}.....eq.1

{{{a[1]=5-9(2a[1])/2}}}

{{{a[1]=5-9a[1]}}}

{{{a[1]+9a[1]=5}}}

{{{10a[1]=5}}}

{{{a[1]=5/10}}}

{{{highlight(a[1]=1/2)}}} -> First term  

go to eq.2

{{{d=2a[1]}}}......... ..eq.2

{{{d=2(1/2)}}}

{{{highlight(d=1)}}}->Common difference


nth term formula:

{{{a[n]=1/2+1(n-1)}}}

first {{{10}}} terms are:

{{{1/2}}}, {{{1/2+1}}}, {{{1/2+2}}}, {{{1/2+3}}},{{{1/2+4}}},{{{1/2+5}}},{{{1/2+6}}}, {{{1/2+7}}},{{{1/2+8}}},{{{1/2+9}}}

or
{{{0.5}}}, {{{1.5}}}, {{{2.5}}}, {{{3.5}}},{{{4.5}}},{{{5.5}}},{{{6.5}}}, {{{7.5}}},{{{8.5}}},{{{9.5}}}

check their sum:

{{{S[10]=10(0.5+9.5)/2}}}
{{{S[10]=5(10)}}}
{{{S[10]=50}}}-> true


The sum of the first {{{100}}} terms:

first find {{{100}}}th term: 

{{{a[1]+99d}}}-> {{{0.5+99*1=99.5}}}

then the sum is:

{{{S[100]=100(0.5+99.5)/2}}}

{{{S[100]=50(100)}}}

{{{S[100]=5000}}}