Heyy guys....
The ratio of the sums of first m and first n terms of an A.P is m^2:n^2.
Show that the ratio of its mth and nth terms is (2m-1) : (2n-1).
Plzz give the answer asap... and no spamming Plzzz
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hope it helps
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Step-by-step explanation:
Sn = n/2(2a+(n-1)d)
Sm = m/2(2a+(m-1)d)
Sm/Sn = [m/2(2a+(m-1)d)] / [n/2(2a+(n-1)d)] = m^2/n^2
[m(2a+(m-1)d)]/[n(2a+(n-1)d] = m^2/n^2
n[2a+(m-1)d] = m[2a+(n-1)d]
2a(n-m) + (mn-n)d = (mn-m)d
2a(n-m) + (m-n)d = 0
2a = d
a = d/2
Show that :
Am/An = a+(m-1)d / a+(n-1)d = (2m-1)/(2n-1)
then
[a+d(m-1)] / [a+d(n-1)]
= [a+2a(m-1)]/[a+2a(n-1)]
= [ 1+2m-2] / [ 1+2n-2]
= (2m-1) / (2n-1)
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