Question

in an A.p given tn=4 d=2,sn= -14 find a and n

Answer 1

a=4;n=1
thnx.........

Answer 2

Answer:

An=a+(n-1)d=4

a+(n-1)2=4 (substitute d=2)

a+2n-2=4

a+2n=6

a=6-2n

Sn=n/2(2a+(n-1)d)

-14=n/2(2(6-2n)+(n-1)2) (substitute a=6-2n,d=2)

-14=n/2(12-4n+2n-2)

-14=n/2(10-2n)

-14=n(5-n)

5n-n^2=-14

n^2-5n-14=0

n^2-7n+2n-14=0

6-2n(n-7)+2(n-7)=0

(n-7)(n+2)=0

n=7,-2

n can't be negative

therefore n=7

a=6-2n (substitute n=7)

a=6-2(7)=6-14=-8