Sum of first n terms ofan Arithmetic progression with first term a and common difference d, is given
by
A) S. = [2a-(n+1)ds) = (2a + (n +1d]
c) S.=[2a+(n-1)d]p) s. = [za-(n-1)d]
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Answer:
Explanation:
The sum of n terms formula is given by,
S
n
=
2
n
[2a+(n−1)d] .. (1)
Given :- S
n
=2n−n
2
... (2)
Comparing equation (1) & (2), we get
2
n
(2a+(n−1)d)=2n−n
2
2an+n
2
d−nd=4n−2n
2
(2a−d)n+dn
2
=4n−2n
2
...(3)
On comparing factors of n
2
in equation (3)
d=−2
On comparing factors of n in equation (3), we have
2a−d=4
2a=4+d
2a=4−2
2a=2
a=1
Therefore, the first term, a=1 and the common difference, d=−2.
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