In an AP: a_n = 4, d = 2, S_n = -14, find n and a. - Vidyadip

Question

In an $AP: a_n = 4, d = 2, S_n = -14$, find n and a.

✓

Answer

Here, $a_n = 4$
$d = 2$
$S_n = -14$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow $ 4 = a + (n - 1)d
$ \Rightarrow $ 4 = a + 2n - 2
$ \Rightarrow $ 4 + 2 = a + 2n
$ \Rightarrow $ 6 = a + 2n
$ \Rightarrow $ a + 2n = 6 ...... (1)
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow - 14 = \frac{n}{2}\left[ {2a + (n - 1)2} \right]$
$ \Rightarrow $ -14 = n[a + (n - 1)]
$ \Rightarrow $ -14 = n (a + n - 1)
$ \Rightarrow $ -14 = n (6 - n - 1) .......From (1), (a + 2n = 6 $ \Rightarrow $ a + n = 6 - n)
$ \Rightarrow $ -14 = n(-n + 5)
$ \Rightarrow $ $-14 = -n^2 + 5n$
$ \Rightarrow $ $n^2 - 7n + 2n - 14 = 0$
$ \Rightarrow $ n(n - 7) + 2(n - 7) = 0
$ \Rightarrow $ (n - 7) (n + 2) = 0
$ \Rightarrow $ n - 7 = 0 or n + 2 = 0
$ \Rightarrow $ n = 7 or n = -2
$ \Rightarrow $ n = - 2 is in admissible as n, being the number of terms, is a natural number.
$\therefore $ n = 7
Putting n = 7 in equation (1), we get
a + 2(7) = 6
$ \Rightarrow $ a + 14 = 6
$ \Rightarrow $ a = 6 - 14
$ \Rightarrow $ a = -8

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