maths
posted by Anonymous .
For an A.P.
sum of 2n terms = 3* sum of n terms
then
sum of 3n terms : sum of n terms = ?

given: sum of 2n terms = 3* sum of n terms
(2n)/2 [2a + (2n1)d ] = 3(n/2) [2a + (n1)d ]
2(2a + 2nd  d) = 3( 2a + nd  d)
4a + 4nd  2d = 6a + 3nd  3d
2a = nd d
2a = nd + d
then
sum of 3n terms : sum of n terms
= sum of 3n terms / sum of n terms
= [ (3n/2) [2a + (3n1)d ] ] / [ (n/2) [2a + (n1)d) ]]
= 3( nd+d + 3ndd) / (nd+d + ndd)
= 3(4nd) / (2nd)
=12nd/(2nd)
= 6/1
or 6 : 1
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