In an arithmetic sequence #t_2=6# and #t_6=16#, how do you find #t_21#?

1 Answer
Jun 19, 2016

Answer:

#t_21=53 1/2#

Explanation:

#color(red)("Preliminaries")#

Consider the place indicator of #n# where #n=1# represents the first term and #n=2# the second term and so on.

If #a# is the first term then

1st term #color(white)(.)->a#
2nd term #->a+k#
3rd term #->a+2k#
4th term #->a+3k#

From this you observe than any term is #a+(n-1)k#

Observe:

n=1 #color(white)(.)->a+(1-1)k = a#
n=2#->a+(2-1)k=a+k#
n=3 #->a+(3-1)k=a+2k#
n=4 #->a+(4-1)k=a+3k#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
so

at #t_2 ->n=2 -> a+k=6#......Equation (1)

at #t_6 ->n=6 ->a+5k=16#.....Equation (2)

Subtract equation (1) from Equation (2)

#=>4k=10#

Thus #color(blue)(k=10/4=5/2)#...........................(3) '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute (3) back into Equation (1)

#color(brown)(a+k=6)color(blue)(->a+5/2=6#

Subtract #5/2# from both sides

#color(blue)(a= 7/2)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
#t_2=6-> 7/2+5/2=6# as required
#t_6=16->7/2+(5xx5/2) = 16# as required
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("Answering the questions")#

#color(brown)(t_21 = a+20k)color(blue)(->7/2+(20xx5/2 )=53 1/2#