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

1 Answer
Mar 15, 2018

Answer:

#a_9 = 105#

Explanation:

Given: arithmetic sequence: #a_10 = 100; a_20 = 50#

An arithmetic sequence has the form: #a_n = a_1+d(n-1)#, where #a_1# is the first number in the sequence, #n# is the number in the sequence and #d# 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#