#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#