The 4th term of an
a.p is 22 and 15th term is 66. find the first term and the common difference.hence find the sum of the serise to 8 terms ?
Answers
regards
cheshta
The 4th term of an a.p is 22 and 15th term is 66. The first term is 10 and the common difference is 4. The sum of the series to 8 terms is 192.
It is given in the question that the sequence is an A.P. So, we know that,
nth term in A.P. is given by:
an = a + (n-1)d
where an is the nth term,
a is the first term,
n is the term number,
d is the common difference.
4th term is represented as
a + (4-1)d = a + 3d
Given, that the 4th term is 22, so,
a + 3d = 22 ...(1)
Similarly, the 15th term is 66, so it can be written as,
a + 14d = 66 ...(2)
Subtracting (2) from (1), we get :
14d - 3d = 66 - 22
⇒ 11d = 44
⇒ d = 4
Replacing the value of 'd' in (1), we get:
a + 3(4) = 22
⇒ a = 22-12 = 10.
Thus, the first term in A.P. 'a' is 10 and their common difference 'd' is 4.
The Sum of n terms in A.P. is given as:
Sn = (n/2)(a + l)
where Sn is the sum of n terms,
l is the last term.
The last term in the sum of 8 terms is a8 i.e. the 8th term.
So, a8 = a + (8-1)d = 10 + 7(4) = 10 + 28 = 38
This is l, so, l = 38.
So, sum of 8 terms is
Sn = (8/2)(10+38) = (4)(48) 192.
So, the sum of the A.P. series of 8 terms is 192.