How do you find the first three terms of the arithmetic series n=31, #a_n=78#, #S_n=1023#?

1 Answer
Aug 26, 2017

First three terms are #-12#, #-9# and #-6#

Explanation:

In an arithmetic series with #a_1# as first term and #d# as common difference, while #n^(th)# term #a_n=a_1+(n-1)xxd#, the sum of series up to #n^(th)# terms is #S_n=n/2(a_1+a_n)=n/2(2a+(n-1)d)#.

Here we have #n=31#, #a_n=a_1+30d=78#

and #S_n=1023=31/2(a_1+78)#

Hence #a_1=1023xx2/31-78=33xx2-78=-12#

and as #78=-12+30xxd# we have #d=(78+12)/30=90/30=3#

Hence, first three terms are #-12#, #-12+3=-9# and #-9+3=-6#