Question

Given
$a _{n} = 4$
,
$d = 2$
,
$s _{n} = - 14$
, find n and a .​

Answer 1

aₙ = a + (n-1) d

4 = a + (n-1) 2

n = (6-a)/2 ......(i)

also, sₙ = n/2 ( 2a + (n-1) d )

-14 = n/2 ( 2a + (n-1)2 )

-28 = n ( 2a + 2n - 2 )

Putting value of n in above equation;

$-28=\frac{6-a}{2}(2a+(\frac{6-a}{2})2-2)\\  -56=6a+24-a^2-49\\a^2+2a-80=0\\a^2+10a-8a-80=0\\(a+10)(a-8)=0\\a=-10\\ a=8$

Putting values of a in (i);

(a=8)

n = 6-8/2

n = -2/2

n = -1, which is not possible as no. of terms cannot be negative.

(a = -10)

n = 6-(-10)/2

n = 8

And, a = -10.