How to find first term and commond difference if the only given are the last term (80) and sum of first 10 terms (530)?

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1 Answer
Somebody N.
Jun 19, 2018

#color(blue)("First term"=26)#

#color(blue)("Common difference"=6)#

Explanation:

This is an arithmetic series.

The #nth# term of an arithmetic series is given by:

#a+(n-1)d \ \ [1]#

Where #bba# is the first term, #bbd# is the common difference and #bbn# is the nth term.

The sum of an arithmetic series is given as:

#S_n=n/2(2a+(n-1)d) \ \ \ [2]#

We are given:

Sum of the first 10 terms is #530# and #n=10#

Using in #[2]#

#530=10/2(2a+(10-1)d)#

#530=10a+45d \ \ \ [3]#

Last term is 80:

Using this in #[1]#

#a+(10-1)d=80 \ \ \ [2]#

#a+9d=80 \ \ \ [4]#

Solving #[3]# and #[4]# simultaneously:

#a+9d=80=>a=80-9d#

in #[3]#

#530=10(80-9d)+45d#

#530-800=-45d=>d=6#

Substituting in #[4]#

#a+9(6)=80#

#a=80-54=26#

First term is:

#color(blue)(26)#

The common difference is:

#color(blue)(60#

Check:

last term:

#26+(10-1)(6)=80#

Sum of first 10 terms:

#10/2(2(26)+(10-1)(6))=10/2(52+54)=5(106)=530#

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