Question 1177181
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The sum of n terms of a certain series is 3n^2 + 10n for all values of n. 
Find the nth term and show that the series is an arithmetical progression.
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<pre>
If the sum of the first n terms of the sequence is given by the formula  {{{S[n]}}} = 3n^2 + 10n, then


{{{a[n]}}} = {{{S[n]}}} - {{{S[n-1]}}} = (3n^2 + 10n) - ( 3(n-1)^2 + 10(n-1) ) =


           = 3n^2 + 10n - (3n^2 - 6n + 3 + 10n - 10) = 6n + 7.



The n-th term of the sequence is  {{{a[n]}}} = 6n + 7.


It is the arithmetic progression with the first term of 13 (at n= 1) and the common difference of 6.


The proof is completed.
</pre>

Solved, answered, explained and completed.