Question

what is the sum of the first 4 terms of the arithmetic sequence in which the 6th term is 8 and the 10th term is 13

Answer 1

so we need to first write what an arithmetic sequence looks like in general
general then we can solve for the terms
a_n = a_n+1 + (n-1)d
this is the general equation
we know that a_6 = 8 and a_10 = 13

so a_7, a_8, a_9 all come in between
we need to add something to 8 4 times to get to 13

(13-8)/4
= 5/4 = 1.25

a_6 = 8
a_7 = 9.25
a_8 = 10.5
a_9 = 11.75
a_10 = 13
so that works

by adding 1.25 to the a_6 term 4 times, we get to 13
so the "d" term is 1.25

we need to back up to get a_1, a_2, a_3 and a_4
might be easier to just subtract d
a_5 = 8 - 1.25 = 6.75

a_4 = 6.75 - 1.25 = 5.5
a_3 = 5.5 - 1.25 = 4.25

a_2 = 4.25 - 1.25 = 3
so the sum is:

1.75 + 3 + 4.25 + 5.5 =)

Answer 2

= 6 + 3 + 5.5 = 9 + 5.5 = 14.5

Answer 3

we could solve for a_n-1
a_n-1 = a_n - (n-1)d
but again
it just backs us up one step at a time

Answer 4

Well the 6th term is a +5d = 8
10th term a+9d = 13
5 = 4d, d = 5/4
a = 8 -25/4 = 7/4

So your nth term = 7/4+(5/4)(n-1) = 1/4(5n+2)
4th term = (1/4)(20+2) = 22/4, get the the sum = 29/2 = 14.5