Give proving of an=Sn-Sn-1
Answers
Answer:
The idea lies at the definition of Sn.
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Suppose a series: a1, a2, a3, a4, a5, ..., a(n-1), an, a(n+1), ....
Here, a1= first term,
an= nth term of the series,
Sn= sum of first n terms of the series,
S(n-1)= sum of first (n-1) terms of the series.
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Therefore, S1= a1 = sum of very first term only
S2= a1+ a2 = sum of first two terms
S3= a1+ a2+ a3 = sum of first three terms
...
S(n-1)= a1+ a2+ a3+ ...+ a(n-1)
Sn= a1+ a2+ a3+ ...+ a(n-1)+ an
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Therefore,
Sn= [ a1+ a2+ a3+ ...+ a(n-1) ]+ an
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Sn= S(n-1) + an
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On substracting 'S(n-1)' on both the sides, we get,
Sn- S(n-1)= an.
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For example =>
assume a series: 4,6,8,3,5,2
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Here n=6= number of last term
a1=4 = first term
a6=2 = an= last term
S1= a1= 4
S2= 4+6
S3= 4+6+8
S4= 4+6+8+3
S5= 4+6+8+3+5
S6= 4+6+8+3+5+2
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Thus, S6= (4+6+8+3+5) +2
S6= S5+ 2
Therefore, S6 - S5=2 -------- ( i )
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Now put n=6 here in equation ( i ), we get,
Sn- S(n-1) = 2 ----- ( ii )
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Here n=6 and (n-1)=5.
Also since, a6= 2= an= last term.
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Hence from ( ii ),
Sn- S(n-1)= an.