How do you find the 9th term of the arithmetic sequence a10 = 100 and a20 = 50?

1 Answer
Mar 15, 2018

a_9 = 105a9=105

Explanation:

Given: arithmetic sequence: a_10 = 100; a_20 = 50a10=100;a20=50

An arithmetic sequence has the form: a_n = a_1+d(n-1)an=a1+d(n1), where a_1a1 is the first number in the sequence, nn is the number in the sequence and dd is the arithmetic difference between consecutive terms: d = a_2 - a_1 = a_3 - a_2 ....

a_10 = a_1 +d*(10-1) = 100
a_10 = a_1 +9d = 100

a_20 = a_1 + d*(20-1) = 50
a_20 = a_1 +19d = 50

Find d:
We have two equations and two unknowns. Use elimination.
Multiply the second equation by -1 and add the two equations:

Equation 1: " "a_1 + 9d = 100
Equation 2 *(-1) = ul(+ -a_1 -19d = -50)
" "-10d = 50

d = -5

Find a_1 by substitution: " "a_1 + 19 * -5 = 50

Simplify: " "a_1 - 95 = 50

a_1 = 50 + 95 = 145

CHECK: values of a_1 and d
#a_10 = 145 + 9*(-5) = 100

#a_20 = 145 + 19*(-5) = 50

Find a_9:
a_9 = a_1 + 8d = 145 + 8(-5) = 105

Does this answer make sense?
Yes, because a_9 is slightly larger than a_10 which is larger than a_20