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