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Method and examples
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Arithmetic Progression |
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Problem 15 of 19 |
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15. If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1)
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Solution
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Solution provided by AtoZmath.com
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This is demo example. Please click on Find button and solution will be displayed in Solution tab (step by step)
| Arithmetic Progression |
15. If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1)
Let a be the first term and d be the common difference
Now,
`S_1= n/2 [ 2a + (n - 1) d ]`
`S_2 = (2n)/2 [ 2a + (2n - 1) d ]`
`S_3 = (3n)/2 [ 2a + (3n - 1) d ]`
Now, `3(S_2 - S_1)`
`= 3 [ (2n)/2 ( 2a + (2n - 1) d ) - n/2 ( 2a + (n - 1) d ) ]`
`= 3 [ n/2 ( 2(2a) - 2a ) + n/2 ( 2(2n - 1) d - (n - 1) d ) ]`
`= 3 [ n/2 ( 2a ) + n/2 ( 4n - 2 - n + 1) d ) ]`
`= 3 [ n/2 ( 2a ) + n/2 ( 3n - 1) d ) ]`
`= (3n)/2 [ 2a + ( 3n - 1) d ]`
`= S_3` (Proved)
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