How do you write the nth term rule for the arithmetic sequence with #a_7=34# and #a_18=122#?

2 Answers
Jul 31, 2016

Answer:

#n^(th)# term of the arithmetic sequence is #8n-22#.

Explanation:

#n^(th)# term of an arithmetic sequence whose first term is #a_1# and common difference is #d# is #a_1+(n-1)d#.

Hence #a_7=a_1+(7-1)xxd=34# i.e. #a_1+6d=34#

and #a_18=a_1+(18-1)xxd=122# i.e. #a_1+17d=122#

Subtracting firt equation from second equation, we get

#11d=122-34=88# or #d=88/11=8#

Hence #a_1+6xx8=34# or #a_1=34-48=-14#

Hence #n^(th)# term of the arithmetic sequence is #-14+(n-1)xx8# or #-14+8n-8=8n-22#.

Answer:

#color(blue)(a_n=8n-22)#

Explanation:

The given data are

#a_7=34# and #a_18=122#

We can set up 2 equations

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

#a_7=a_1+(7-1)*d#

#34=a_1+6*d" "#first equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#a_n=a_1+(n-1)*d#

#a_18=a_1+(18-1)*d#

#122=a_1+17*d" "#second equation

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

By method of elimination using subtraction, let us use first and second equations

#34=a_1+6*d" "#first equation

#122=a_1+17*d" "#second equation

By subtraction, we have the result

#88=0+11d#

#d=88/11=8#

Solving now for #a_1# using the first equation and #d=8#

#34=a_1+6*d" "#first equation

#34=a_1+6*8" "#

#34=a_1+48#

#a_1=-14#

We can write the #nth # term rule now

#a_n=-14+8*(n-1)

#a_n=-14-8+8n#

#color(blue)(a_n=8n-22)#

God bless....I hope the explanation is useful.