Question

Let sn denote the sum of first n terms of an

a.p..if s2n = 3sn, then the ratio s3n / 5n is equal to

Answer 1

Sn=n/2(2a+(n-1)d )
3Sn=3n/2(2a+(n-1)d ).......1
S2n=n(2a+(2n-1)d )
B/Q
2a+(2n-1)d =3/2*2a+3/2(n-1)d
2a+2nd-d =3a+3/2nd-3/2 d
after simplifying
2a=(n+1)d
Now
S3n=3n/2 (2a+(3n-1)d )
S3n=3n/2((n+1)d + (3n-1)d )
=3n/2 (nd+d+3nd-d)
= 3n/2*4nd
=6nsquare d
therefore
S3n/5n=6n square d/5n
=6/5 nd

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