Question

Tn=4 , d=2 and Sn =-14 . Find n and a.

Answer 1

..,....... ......... ................

Answer 2

In an AP let:

a be the first term

d be the common difference

We know nth term is a+(n-1)d

Given:

nth term=4

So:

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

Now we also know that:

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

Given n=-14

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

From(1) we get:

==) -14=n/2*[a+4]

==) -28=n*[a+4]

==) -28/n=a+4

==) a=-28/n-4.........(2)

In the first equation:

a+(n-1)d=4

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

==)a+2n-2=4

==)a+2n=6

==)From 2 we get:-

==)-28/n-4+2n=6

==)-28/n+2n-10=0

Multiplying by n:

==)-28+2n²-10n=0

==)2n²-10n-28=0

==)2n²-14n+4n-28=0

==)2n(n-7)+4(n-7)=0

==)(2n+4)(n-7)=0

Either 2n+4=0 or n-7=0

So:

2n=-4

n=-2 or:

n=7

But n cannot be negative as number of terms cannot be negative

So n=7.

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

So:

==) a+2(7-1)=4

==) a+2*6=4

==) a+12=4

==)a=4-12

==) a=-8

The value of a is -8 and n is 7.

Hope it helps.