Gmat_mission wrote:In the sequence a_1, a_2, . . . , a_n each term after the first term is equal to the preceding term plus a constant c. $$a_1+a_{11}+a_{21}=99.$$ What is the value of $$a_3+a_{19}=?$$ A. 66
B. 44
C. 33
D. 22
E. 11
For any evenly spaced set:
average = median.
sum = (count)(average).
a� + a�� + a₂� = 99.
In the given sequence, the difference between one term and the next = c.
Implication:
{a�, a��, a₂�} constitutes an evenly spaced set because the difference between one term and the next = 10c.
Thus, the median of the set -- in other words, the value of a�� -- is equal to the average of the 3 terms:
a�� = sum/count = 99/3 = 33.
a�� is also the median -- and thus the average -- of {a₃, a�₉}.
Thus, the sum of these 2 terms = (count)(average) = (2)(33) = 66.
The correct answer is
A.
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