How do you find the first term and common difference that has a sum of its first 10 terms equal to 250 and whose 6th term is 32?

1 Answer
Sep 13, 2015

Answer:

The first term is #-38# (negative #38#).
The common difference is #14#.

Explanation:

So, we are talking about arithmetic progression, that is (this is the definition) a sequence of numbers, starting with some first one, #a#, and each consecutive one differed from the previous by the common difference #d#.

That is, we are talking about a sequence
#a, a+d, a+2*d, a+3*d,..., a+N*d,...#

Assuming you don't remember the formula for this, let's derive a formula for a sum of the first #N# terms of this sequence:
#S_N = [a] + [a+d] + [a+2*d] +...+ [a+(N-2)*d] + [a+(N-1)*d]#

Since the sum does not change if we change the order of summation, we can summarize it in opposite order:
#S_N = [a+(N-1)*d] + [a+(N-2)*d] +...+ [a+2*d] + [a+d] + [a]#

Sum both #S_N# adding first term to first term, second term to second term etc., #(N-1)^(th)# term to corresponding #(N-1)^(th)# term, getting:

#S_N+S_N = #
#= {[a] + [a+(N-1)d]} +#
#+ {[a+d] + [a+(N-2)d]} +...#
#...+ {[a+(N-2)d] + [a+d]} +#
#+ {[a+(N-1)d] + [a]} #

Notice, that the results of summation in each #{...}# is the same, #2a+(N-1)*d#.

Therefore,
#2S_N= {2a+(N-1)*d}*N#

and
#S_N = a*N + [(N-1)*N*d]/2#

The problem states that for #N=10# #S_10=250#.
Therefore, we have one equation:
(Eq. 1) #250 = 10a+45d#

Since #6^(th)# term is #32#, we have another equation:
#a+(6-1)d=32# or
(Eq. 2) #a+5d=32#

We have a system of two equations, Eq. 1 and Eq. 2, with two unknowns #a# and #d#.
It is easy to solve it using a substitution method.

From equation Eq. 2:
#a = 32-5d#

Substitute this value for #a# into equation Eq. 1:
#250 = 10(32-5d)+45d#
or
#250 = 320-50d+45d#,
from which:
#5d=70# and
#d=14#
That is the common difference of our sequence.

Back to the unknown #a=32-5d#:
#a=32-5*14=-38#

Checking (ALWAYS RECOMMENDED)

The first 10 members of this sequence that starts with #-38# and adds #14# to each new member are
#-38, -24, -10, 4, 18, 32, 46, 60, 74, 88#.
Their sum is indeed #250# and their #6^(th)# term is indeed #32#.
Checking confirms the validity of the answer.