How do you write a rule for the nth term of the arithmetic sequence given #a_20=240, a_15=170#?

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Answer 1 · 2017-07-20T06:43:36

#n^(th)# term #a_n=14n-40#

#n^(th)# term #a_n# of an arithmetic sequence, whose first term is #a_1# and common difference is #d# is given by

#a_n=a_1+(n-1)d#

Hence #a_20=a_1+19d=240# ..........(A)

and #a_15=a_1+14d=170# ..........(B)

Subtracting (B) from (A), #5d=70# i.e. #d=70/5=14#

and hence putting this in (B), we get

#a_1+14xx14=170#

or #a_1=170-14xx14=170-196=-26#

Hence #a_n=-26+(n-1)xx14#

#=-26+14n-14=14n-40#

Answer 2 · 2017-07-20T06:45:07

#a_"n"=-26+14(n-1)#

#ω=(a_20-a_15)/5=(240-170)/5=70/5=14#

#a_1=a_15-14ω=170-196=-26#

So for the #n^{th}# term we know :

#a_"n"=a_1+(n-1)ω=-26+14(n-1)#

To check out answer let's calculate #a_20#

#a_20=-26+14*19=-26+266=240#

Which is true.