Vincen wrote: ↑

Thu Sep 03, 2020 6:07 am

Let \(T\) be a sequence of the form \(a_n=a_1+d\cdot (n-1).\) If \(a_3=17\) and \(a_{19}=65,\) find \(a_{10}.\)

A. 37
B. 38
C. 39
D. 40
E. 41

Answer: B

Solution:

The given formula is actually a formula for terms of an arithmetic sequence, i.e., a sequence whose terms are generated by adding or subtracting the same constant (called common difference, or the value of d in the formula.) from one term to the next. Since a_3 = 17 and a_19 = 65, the difference between the two terms is 65 - 17 = 48 and the difference between the indices of the two terms is 19 - 3 = 16. It follows that the common difference is d = 48/16 = 3.

To find a_10, we can also use a slightly different formula: a_n = a_m + d(n - m). Letting n = 10 and m = 3, we have:

a_10 = a_3 + 3(10 - 3) = 17 + 21 = 38

[Note: We can also let m = 19 (where n is still 10) and obtain:

a_10 = a_19 + 3(10 - 19) = 65 - 27 = 38]

Answer: B