in an a. p an =4, d=2, sn= _14 find n and a
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a
n
=a+(n−1)d
4=a+(n−1)2
4=a+2n−2
6=a+2n
∴a=6−2n...(1)
S
n
=
2
n
[2a+(n−1)d]
−14=
2
n
[2(6−2n)+(n−1)2]
−28=n[12−4n+2n−2]
−28=12n−2n
2
−2n
−28=10n−2n
2
−14=5n−n
2
Solving the above equation, we get,
(n−7)(n+2)=0
n=7,−2
∴n=7
a=6−2n From (1)
a=6−2(7)=−8
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