The sum of the first 10 terms of an arithmetic sequence is 155. What is the sum of the 5th and the 6th terms? If the 2nd term is 5, what is 9th term? What is the common difference of the sequence?
Answers
I think this is wrong question
The sum of the 5th and the 6th terms is 31. The 9th term is 26. Common difference of the sequence is 3.
We know, the nth term in A.P is given by:
an = [a + (n-1)d],
where a is the first term,
d is the common difference between consecutive 2 terms.
And sum of first n terms is given as:
S = (n/2)[2a + (n-1)d]
Given 2nd term is 5.
So, according to the first relation,
a + (2-1)d = 5
⇒a + d = 5
⇒ a = 5 - d ...(1)
Again, sum of first 10 numbers is 155.
So, according to the second equation,
(10/2)[2a + (10-1)d] = 155
⇒ 2a + 9d = 155/5 = 31 ...(2)
Putting the value of a from (1) in (2),
2(5-d) + 9d = 31
⇒ 10 - 2d + 9d = 31
⇒ 7d = 31 - 10 = 21
⇒ d = 21/7 = 3
So, common difference = 3.
As a = 5 - d
So, a = 5 - 3 = 2
First term is 2.
5th term is :
2 + (5-1)(3) = 2 + 12 = 14
6th term is :
2 + (6-1)(3) = 2 + 15 = 17
So, sum of 5th and 6th term is (14 + 17) = 31
9th term is
2 + (9-1)(3) = 2 + 24 = 26.