What is the 32nd term of the arithmetic sequence where a1 = –32 and a9 = –120?

2 Answers
Nov 8, 2015

Answer:

#A_32 = -373#

Explanation:

#A_n = A_1 + d(n - 1)#

#A_1 = -32#

#A_9 = -120#

#=> A_9 = -32 + d(9 - 1)#

#=> -120 = -32 + d(8)#

#=> -88 = 8d#

#=> d = -11#


#A_32 = -32 + (-11)(32 - 1)#

#=> A_32 = -32 + (-11)(31)#

#=> A_32 = -32 + -341#

#=> A_32 = -373#

Jul 29, 2017

Answer:

#a_32 = -373#

Explanation:

Any term in an arithmetic sequence can be found from the formula

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

So if can find the value of #a_1 and d# we can find any term,

In this case we are told that #a_1 = -32#

For the #9th# term, #n= 9# and #a_ = -120#

Use these values in the general formula.

#color(white)(......)color(blue)(a_ncolor(white)(...) = color(white)(...)a_1 color(white)(.......)+(n-1)d)#
#color(white)(......)uarrcolor(white)(..........)uarrcolor(white)(...............)uarrcolor(white)(...)uarr#
#" "-120color(white)(....)-32color(white)(..............)8color(white)(....)?#

This gives the equation:

#-120 = -32 +8d#

#-120+32 = 8d#

#-88 = 8d#

#d = -11#

Now the formula for the general term becomes:

#color(blue)(a_n = -32 + (n-1)(-11))" "larr# simplify

#color(blue)(a_n = -32 + -11n+11)#

#color(blue)(a_n = -11n -21)#

Find the #32nd# term (#n =32#)

#a_32 = -11(32)-21#

#a_32 = -352-21#

#a_32 = -373#