Given that $a_1, a_2 , a_3, \cdots, a_{21}$ are in an arithmetic progression and the value of
\[\sum_{i = 1}^{21} a_i = 693\]
Find the value of
\[\sum_{r = 0}^{10} a_{2r + 1}\]
+128406 In th first, we have that the sum of an arithemetic series is given by
(first term + last term) (n /2)
So
(first term + last term) (21 /2 = 693
first term + last term = 693 ( 2) / 21 = 66
And in the second summation.....we are summing these terms of the first series :
Term # + Term #
1 + 21 = 66
3 + 19 = 66
5 + 17 = 66
7 + 15 = 66
9 + 13 = 66
11 = 66/2 = 33
So the sum = 66(5) + 33 = 363
