The sum of the first 40 terms of an A.P. is 3980 while the sum of the first 20 terms of the same A.P. is 990. The first term of the A.P. is 2. find The ratio of the 40 th term to the 20 th term of the A.P.
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let 40th term= a(40)=x
s(n)=n[{a(1)+a(40)}/2]
i.e. 3980=40[{2+x}/2]
i.e. 3980=20(2+x)
i.e. 3980=40+20x
i.e. 20x=3940
i.e. x=3940/20=197
let 20th term= a(20)=y
s(n)=n[{a(1)+a(20)}/2]
i.e. 990=20[{2+y}/2]
i.e. 990=10(2+y)
i.e. 990=20+10y
i.e. 10y=970
i.e. y=970/10=97
i.e. the ratio of 40th term to 20th term=197/97
s(n)=n[{a(1)+a(40)}/2]
i.e. 3980=40[{2+x}/2]
i.e. 3980=20(2+x)
i.e. 3980=40+20x
i.e. 20x=3940
i.e. x=3940/20=197
let 20th term= a(20)=y
s(n)=n[{a(1)+a(20)}/2]
i.e. 990=20[{2+y}/2]
i.e. 990=10(2+y)
i.e. 990=20+10y
i.e. 10y=970
i.e. y=970/10=97
i.e. the ratio of 40th term to 20th term=197/97
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