What is the 17th term in the arithmetic sequence in which #a_6# is 101 and #a_9# is 83?

1 Answer
Jun 14, 2016

Answer:

#17^(th)# term is #35#

Explanation:

#n^(th)# term of an arithmetic sequence #{a,a+d,a+2d,........}#, whose first term is #a# and difference between a term and its preceeding term is #d#, is #a+(n-1)d#.

As #6^(th)# term is #101#, we have #a+5d=101# and .......(1)

as #9^(th)# term is #83#, we have #a+8d=83#. ......(2)

Subtracting (1) from (2), we get #3d=-18# i.e, #d=-6#

Putting this in (1), we get #a-30=101# or #a=131#

Hence, #17^(th)# term is #131+16xx(-6)=131-96=35#