Question

Given the sequence 8, 12, 16, 20, 24, what is the sum of the 31st and 19th terms?

Answer 1

Answer:

First term (a) =8

Common difference (d)= t2-t1

=12-8

=4

Now, sum of first 31th term (tn31) =n/2{2a+(n-1)d}

= 31/2{2×8+(31-1)4}

=31/2{16+(30×4)

=31/2(16+120)

=31/2×126

=31×63

Step-by-step explanation:

Similarly use 19 as (n) for the 19th term