Question

If the ratio of sum of 1st m and n terms of an AP is m square:n square. Show that the ratio of its mth and nth term is (2m+1):(2n-1).

Answer 1

Sm= m/2[2a+[m-1]d]= m2

Sn= n/2[2a+[n-1]d]= n2

so n2 will cut n n similarly m2 will cut m

Then, 2an+[mn-n]d= 2am+[mn-m]d

swift all similar terms one side

2an-2am=[mn-m-mn+n]d

2a[n-m]=[n-m]d

2a=d

am/an=a+[m-1]d/a+[n-1]d

replace d with 2a

a+2am-2a=2am-a

a+2an-2a=2an-a

no take 'a' common

n here is the result

i.e 2m-1/2n-1

[you have written ques wrong]