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The sum of the first n terms of an arithmetic sequence is given by Sn=3n^2 - 2n. What...
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You need to remember what the common difference of an arithmetic progression means.
`a_(n) - a_(n-1) = d`
The problem provides the information that the sum of n terms of arithmetic progression is `3n^2-2n` such that:
`a_1+a_2+....+a_(n-1)+a_n = 3n^2-2n `
Hence, you may evaluate the sum of n-1 terms of arithmetic progression such that:
`a_1+a_2+....+a_(n-1) = 3(n-1)^2-2(n-1)`
Hence, subtracting the sum of n-1 terms from the sum of n terms yields a_n such that:
`S_n - S_(n-1) = a_1+a_2+....+a_(n-1)+a_n - a_1-a_2+....-a_(n-1)`
Reducing like terms yields `a_n` :
`a_n = S_n - S_(n-1) `
`a_n = 3n^2 - 2n - 3(n-1)^2 + 2(n-1)`
You need to open the brackets such that:
`a_n = 3n^2 - 2n - 3n^2 + 6n - 3 + 2n - 2`
Reducin lke terms yields:
`a_n = 6n - 5`
You may evaluate the term `a_(n-1) ` such that:
`a_(n-1) = 6(n-1) - 5`
`a_(n-1) = 6n - 6 - 5`
`a_(n-1) = 6n - 11`
You may evaluate the common difference such that:
`a_(n) - a_(n-1) = 6n - 5 - 6n + 11`
`a_(n) - a_(n-1) = 6`
Hence, evaluating the common difference of arithmetic progression under given conditions yields d = 6.
Posted by sciencesolve on May 4, 2012 at 7:37 AM (Answer #1)
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