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November 28, 2014

November 28, 2014

Posted by **Ama** on Wednesday, November 21, 2012 at 10:49pm.

2, 5, 8, 11, . . .

- precalculus -
**Bosnian**, Thursday, November 22, 2012 at 2:55amThe sum of the members of a arithmetic progression :

Sn = ( n / 2 ) * [ 2 a1 + ( n - 1 ) * d ]

d = common difference of successive members of arithmetic progression

In this case :

S1 = ( 1 / 2 ) * [ 2 * 2 + ( 1 - 1 ) * 3 ]

S1 = ( 1 / 2 ) * (4 + 0 * 3 )

S1 = ( 1 / 2 ) * 4 = 2

S2 = ( 2 / 2 ) * [ 2 * 2 + ( 2 - 1 ) * 3 ]

S2 = 1 * ( 4 + 1 * 3 )

S2 = 4 + 3 = 7

S3 = ( 3 / 2 ) * [ 2 * 2 + ( 3 - 1 ) * 3 ]

S3 = ( 3 / 2 ) * ( 4 + 2 * 3 )

S3 = ( 3 / 2 ) * ( 4 + 6 )

S3 = ( 3 / 2 ) * 10

S3 = 30 / 2 = 15

S4 = ( 4 / 2 ) * [ 2 * 2 + ( 4 - 1 ) * 3 ]

S4 = 2 * ( 4 + 3 * 3 )

S4 = 2 * ( 4 + 9 )

S4 = 2 * 13

S4 = 26

S5 = ( 5 / 2 ) * [ 2 * 2 + ( 5 - 1 ) * 3 ]

S5 = ( 5 / 2 ) * ( 4 + 4 * 3 )

S5 = ( 5 / 2 ) * ( 4 + 12 )

S5 = ( 5 / 2 ) * 16

S5 = 80 / 2 = 40

S6 = ( 6 / 2 ) * [ 2 * 2 + ( 6 - 1 ) * 3 ]

S6 = 3 * ( 4 + 5 * 3 )

S6 = 3 * ( 4 + 15 )

S6 = 3 * 19

S6 = 57

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