What is the 20th term of the arithmetic sequence 2, –4, –10, … ?

3 Answers
Mar 5, 2018

Answer:

#-112#

Explanation:

the nth term of an AP

#a,a+d,a+2d,..#

is #n^(th)=a+(n-1)d#

here we have

#2,-4,-10,...#

#a=2#

and it is seen that #d=-6#

#:. 20^(th)=2+(20-1)(-6)#

#=2+19xx-6=2-114#

#=-112#

Mar 5, 2018

Answer:

#20^(th)# term #=-112#

Explanation:

If the sequence is denoted by the series #a_i#
then #a_i=a_(i-1)-6#

Setting #a_0=8#
so that the first term is #a_1=2# (as given)

we have #a_n=a_0-(n * 6)#

For #n=20#
#color(white)("XXX")a_20 = 8- 20*6=8-120=-112#

Mar 5, 2018

Answer:

#T_20 = -112#

Explanation:

The terms in the sequence #2, -4, -10 ...# all differ by #-6#

You can keep writing the terms until you get to the 20th, but that is not a very mathematical way of doing it and would be VERY time-consuming if you were asked for the 200th term for example.

Find the rule for the sequence... called #T_n#

If the difference is #-6#, then the rule starts with #T_n = -6n#

You need to adjust the rule to start at the correct number,
#n# starts from #1, 2, 3, 4 ...# for each successive term.

When #n=1, " "-6(1) = -6#

But the first term must be #2# therefore add #8#

#T_n = -6n+8#

Check if #n=2#
#T_2 = -6(2)+8 = -12+8 =-4#

Check if #n=3#
#T_3 = -6(3)+8 = -18+8 =-10#

Now find the 20th term:

#T_20 = -6(20)+8#

#=-120+8 = -112#