Q : 3      In an AP:

           (viii) given \small a_n=4,d=2,S_n=-14,  find \small n and \small a .

Answers (1)

It is given that
\small a_n=4,d=2,S_n=-14,
Now, we know that
a_n = a+(n-1)d
4 = a+(n-1)2
a+2n = 6\Rightarrow a = 6-2n \ \ \ \ \ \ \ \ \ \ \ \ \ -(i)

Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow -14 = \frac{n}{2}\left \{ 2\times(a) +(n-1)2\right \}
\Rightarrow -28 = n\left \{ 2(6-2n)+2n-2 \right \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))
\Rightarrow -28 = n\left \{ 10-2n \right \}
\Rightarrow 2n^2-10n-28=0
\Rightarrow 2(n^2-5n-14)=0
\Rightarrow n^2-7n+2n-14=0
\Rightarrow(n+2)(n-7)=0
\Rightarrow n = -2 \ \ and \ \ n = 7
Value of n cannot be negative so the only the value of n is 7
Now, put this value in (i) we will get
a = -8
Therefore, value of n and a are 7 and -8  respectively

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