Question

Ap it is given that tn=4 d=2 sn=14 find n and a

Answer 1

=a+(n-1)d=4

or, a+(n-1)2=4

or, a+2n-2=4

or, a+2n=4+2

or, a+2n=6

or, a=6-2n

=n/2[2a+(n-1)d]=-14

or, n/2[2a+(n-1)2]=-14

or, an+n(n-1)=-14

or, (6-2n)n+n²-n=-14

or, 6n-2n²+n²-n=-14

or, -n²+5n=-14

or, n²-5n=14

or, n²-5n-14=0

or, n²-7n+2n-14=0

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

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

either, n-7=0

or, n=7

or, n+2=0

or, n=-2

∵, n can not be negative ;

∴, n=7

∴, a=6-(2×7)

=6-14

=-8

∴, n=7 and a=-8 Ans.

HOPE THIS HEPLS..................!!!!!

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