Question

in an ap an=4 , d=2 , sn= -14 find n and a

Answer 1

aa=a+(n−1)d

4=a+(n−1)2

4=a+2n−2

6=a+2n

∴a=6−2n...(1)

Sn =2n [2a+(n−1)d]−14

=2n [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