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The third and ninth term of an A.P. are $4$ and $-8$ respectively.
(i) Which term of the A.P. is zero?
(ii) Find the value of $n$ if $S_n = -36$.
(i) Which term of the A.P. is zero?
(ii) Find the value of $n$ if $S_n = -36$.
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Solution: (i)
$a + 2d = 4$ ..... (i) (1/2 Mark)
$a + 8d = -8$ ..... (ii) (1/2 Mark)
solving (i) and (ii) to get $d = -2$, $a = 8$ (1/2 + 1/2 Mark)
$a_n = 0 \Rightarrow 8 + (n - 1) (-2) = 0$ (1 Mark)
$\Rightarrow n = 5$ i.e. $5^{th}$ term is zero.
(ii) $S_n = -36 \Rightarrow \frac{n}{2}[16 + (n - 1) (-2)] = -36$ (1 Mark)
$\Rightarrow n^2 - 9n - 36 = 0$
$\Rightarrow (n - 12) (n + 3) = 0 \Rightarrow n = 12$ (rejecting $n = -3$) (1/2 + 1/2 Mark)
$a + 2d = 4$ ..... (i) (1/2 Mark)
$a + 8d = -8$ ..... (ii) (1/2 Mark)
solving (i) and (ii) to get $d = -2$, $a = 8$ (1/2 + 1/2 Mark)
$a_n = 0 \Rightarrow 8 + (n - 1) (-2) = 0$ (1 Mark)
$\Rightarrow n = 5$ i.e. $5^{th}$ term is zero.
(ii) $S_n = -36 \Rightarrow \frac{n}{2}[16 + (n - 1) (-2)] = -36$ (1 Mark)
$\Rightarrow n^2 - 9n - 36 = 0$
$\Rightarrow (n - 12) (n + 3) = 0 \Rightarrow n = 12$ (rejecting $n = -3$) (1/2 + 1/2 Mark)