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December 18, 2014

December 18, 2014

Posted by **Anonymous** on Tuesday, January 8, 2013 at 3:41am.

sum of 2n terms = 3* sum of n terms

then

sum of 3n terms : sum of n terms = ?

- maths -
**Reiny**, Tuesday, January 8, 2013 at 8:23amgiven: sum of 2n terms = 3* sum of n terms

(2n)/2 [2a + (2n-1)d ] = 3(n/2) [2a + (n-1)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 + (3n-1)d ] ] / [ (n/2) [2a + (n-1)d) ]]

= 3( nd+d + 3nd-d) / (nd+d + nd-d)

= 3(4nd) / (2nd)

=12nd/(2nd)

= 6/1

or 6 : 1

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