the ratio of the sums of first m and first n terms of an arithmetic series is m^2: n^2. show that the ratio of the mth and nth terms is (2m_1):2n_1).
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we convert this in standard formula ,
e.g Sn=n/2{2a+(n-1) d}
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Sm/Sn=m^2/n^2
=m/2 {2.1+(m-1) 2}/n/2 {2.1+(n-1) 2}
now you see
for Sm and Sn
a=1 and d=2
now,
Tm/Tn={a+(n-1) d}/{a+(n-1) d}
=(1+2m-2)/(1+2n-2)=(2m-1)/(2n-1)
e.g Sn=n/2{2a+(n-1) d}
===============================
Sm/Sn=m^2/n^2
=m/2 {2.1+(m-1) 2}/n/2 {2.1+(n-1) 2}
now you see
for Sm and Sn
a=1 and d=2
now,
Tm/Tn={a+(n-1) d}/{a+(n-1) d}
=(1+2m-2)/(1+2n-2)=(2m-1)/(2n-1)
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