Question

given an=4,d=2,sn=[-14],find n and a​.

Answer 1

Answer:

an=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.

Answer 2

Answer:

given;

an=4

d=2

sn= -14

n=?

a=?

we know that;

$an = a + (n - 1).d \\ = 4 = a + (n - 1) \times2 \\ = 4 = a + 2n - 2 \\ = 4 + 2 = a + 2n \\ = 6 \div 2 = a + n \\ = 3 - n = a$