Question

If the sum of the first eighteen terms of an arithmetic progression is three times the sum of the first ten terms while the sum of the fifth and sixteenth terms is 44. Find the sum of the terms from seventh to nineteenth terms.

Answer 1

Answer:

111

Step-by-step explanation:

Let the first term be a  and common difference be d

So 18th term = a+17d

Sum to 18 terms = 9(2a+17d)

Similarly sum to 10 terms = 5(2a+9d)

9(2a+17d) = 3*5(2a+9d)

18a+153d = 30a+135d

12a =1 8d

a=1.5d -------- (1)

Fifth term =  a+4d

16th term = a+15d

Sum = 2a+19d = 44

Substituting a=1.5d

2(1.5d)+19d = 44

22d =44

d=2

a=3

17th term  = 3+16*2= 35

18th term = 3+17*2 = 37

19th term = 3+18*2 = 39

Sum = 35+37+39=111