Evaluate each arithmetic series described. 5) 9+15+21+27…, n=9

Question

Sequences and series assignment

Answer 1

The 9th term is 9+8×6=57

Sum=9+15+21+27+...+51+57.

This can be written 9+(9+6)+(9+12)+(9+18)+...+(9+42)+(9+48). We have nine 9s.

And this is simplified to 9×9+6+12+18+...+42+48=

81+6(1+2+3+...+7+8).

Look at the sum in parentheses. This can be simplified by pairing numbers:

(1+8)+(2+7)+(3+6)+(4+5)=9+9+9+9=36. There are 4 9s because we have 8/2=4 pairs, and each pair sums to 9.

So the sum of the series is 81+6×36=81+216=297. 

If you prefer formulas, let a be the first term, d the common difference and n the number of terms. We have ½(n-1) pairs and each pair sums to 1+(n-1)=n.

The sum S_n=an+½dn(n-1). Let's see if this is correct using n=a=9, d=6:

S₉=81+½(6×9×8)=81+216=297. So the formula is correct.