How do you write a rule for the nth term of the arithmetic sequence given #a_7=6/7, a_9=2/3#?

1 Answer
Feb 27, 2017

Answer:

#a_n= (32-2n)/21#

Explanation:

#a_8# would be #1/2 (a_7 +a_9)#

=#1/2(6/7 +2/3)= 16/21#

Common difference 'd' would thus be#(16/21-6/7)= -2/21#

If the first term is say 'x, then 7th term #a_7= x+ (7-1)(-2/21)#

Thus #6/7= x-4/7#, Or #x=10/7#

nth term expresssion #a_n= a_1 +(n-1)d# can now be written as #a_n= 10/7 +(n-1)(-2/21)#

If required, this can be put in a compact form as

#a_n= (-2n)/21+32/21= (32-2n)/21#