Answer:
a = 5.5, d = 11/3
Step-by-step explanation:
An arithmetic progression is a sequence of number such that the difference between consecutive numbers is a constant. It is given by:
[tex]a_n=a+(n-1)d\\a_n \ is\ the \ nth\ term, d\ is\ the\ common\ difference\ and\ n \ is\ the \ number\ of\ terms[/tex]
The sum of fifth and sixth terms is 44
Therefore:
a + 4d + a + 5d = 44
2a + 9d = 44 (1)
The sum of the first n terms is:
[tex]S=\frac{n}{2} (2a+(n-1)d)\\Therefore\ the\ sum\ of\ the\ first\ 18\ term\ is:\\\\S=\frac{18}{2} (2a+(18-1)d) = 9(2a+17d)=18a+153d\\\\The\ sum\ of\ the\ first\ 10\ term\ is:\\\\S=\frac{10}{2} (2a+(10-1)d) = 5(2a+9d)=10a+45d[/tex]
Since the sum of the first eighteen terms is thrice the sum of the first ten terms:
18a + 153d = 3(10a + 45d)
18 a + 153d = 30a + 135d
18a - 30a + 153d - 135d = 0
-12a + 18d = 0 (2)
Multiply equation 1 by 6 and add to equation 2:
72d = 264
d = 11/3
Putting d = 11/3 in eqn 2
-12a + 18(11/3) = 0
-12a + 66 = 0
12a = 66
a = 5.5
