How do you find the first three terms of the arithmetic series #a_1=11#, #a_n=110#, #S_n=726#?

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Answer 1 · 2016-11-08T18:49:10

First three terms: #11, 20, 29#.

Write a systems of equations.

#t_n = a + (n - 1)d#

#s_n = n/2(2a + (n - 1)d)#

So,

#110 = 11 + (n - 1)d#
#726 = n/2(22 + (n - 1)d)#

Simplify before using substitution.

#726 = n/2(22 + nd - d)#

#726 = (22n + n^2d - nd)/2#

#1452 - 22n = n^2d - nd#

#1452 - 22n = d(n^2 - n)#

#(1452 - 22n)/(n^2 - n) = d#

#110 = 11 + (n - 1)((1452 - 22n)/(n^2 - n))#

#99 = (n - 1)(1452 - 22n)/(n(n - 1))#

#99 = (1452 - 22n)/n#

#99n = 1452 - 22n#

#121n = 1452#

#n = 12#

Now, we can substitute into equation #1# to find the common difference.

#110 = 11 + (12 - 1)d#

#99 = 11d#

#d = 9#

The first 3 terms are #11, 20 and 29#.

Hopefully this helps!