How do you find the formula for #a_n# for the arithmetic sequence #a_3=94, a_6=85#?

2 Answers
May 5, 2018

Answer:

#n# th term is #a_n= 100 -3(n-1)#

Explanation:

#3# rd term #a_3= a_1+2 d =94; (1); a_1 and d # are first term

and common difference of A.P sequence.

#6# th term #a_6= a_1+5 d =85 ; (2); . Subtracting equation (1)

from equation (2) we get, #3 d = -9 :. d= -9/3 =-3# Putting

#d=-3# in equation (1) we get, # a_1+2 * (-3) =94# or

#a_1=94+6=100 :. # first term #a_1=100# Therefore,

#n# th term is #a_n= a_1+(n-1) d or a_n= 100 -3(n-1)# [Ans]

May 5, 2018

Answer:

#a_n=103-3n#

Explanation:

Since this is an arithmetic sequence:
for some constant #k#
#color(white)("XXX")a_6=color(blue)(a_5)+k#
#color(white)("XXX"a_6)=color(blue)(""(color(lime)(a_4)+k))+k#
#color(white)("XXX"a_6)=color(blue)(""(color(lime)(""(a_3+k))+k))+k#
#color(white)("XXX"a_6)=a_3+3k#

since #a_6=85# and #a_3=64#
#color(white)("XXX")3k = 85-94=-9#
#color(white)("XXX")rArr k=-3#

Similarly
#color(white)("XXX")a_1=a_0+k#
#color(white)("XXX")a_2=a_0+2k#
#color(white)("XXX")a_3=a_0+3k#
and
since #a_3=94# and #k=-3#
#color(white)("XXX")94=a_0+3 * (-3)=a_0-9#

#color(white)("XXX")rArr a_0=103#

The general arithmetic formula is
#color(white)("XXX")a_n=a_0+ k * n#
So, for this specific case,
#color(white)("XXX")a_n=103+(-3)n#