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# In an AP: (viii) given an = 4, d = 2, Sn = -14, find n and a .

Q : 3      In an AP:

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

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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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