Question
In an AP:$ a = 8, a_n= 62, S_n= 210,$ find $n$ and $d$.
✓
Answer
Here, $a = 8$
$a_n= 62$
$S_n= 210$
We know that
$a_n= a + (n - 1)d$
$ \Rightarrow 62 = 8 + (n - 1)d$
$ \Rightarrow 62 - 8 = (n - 1)d$
$ \Rightarrow 54 = (n - 1)d$
$ \Rightarrow (n - 1)d = 54 ........ (1) $
Again we know that
${S_n} = \frac{n}{2}\left[ {2n + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {2(8) + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {16 + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {16 + 54} \right]$$ .........$ Using $(1)$
$ \Rightarrow 210 = \frac{n}{2}(70)$
$ \Rightarrow 210 = 35n$
$ \Rightarrow n = \frac{{210}}{{35}}$
$ \Rightarrow n = 6$
Putting $n = 6$ in equation $(1)$, we get
$(6 - 1)d = 54$
$ \Rightarrow 5d = 54$
$ \Rightarrow d = \frac{{54}}{5}$
$a_n= 62$
$S_n= 210$
We know that
$a_n= a + (n - 1)d$
$ \Rightarrow 62 = 8 + (n - 1)d$
$ \Rightarrow 62 - 8 = (n - 1)d$
$ \Rightarrow 54 = (n - 1)d$
$ \Rightarrow (n - 1)d = 54 ........ (1) $
Again we know that
${S_n} = \frac{n}{2}\left[ {2n + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {2(8) + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {16 + (n - 1)d} \right]$
$ \Rightarrow 210 = \frac{n}{2}\left[ {16 + 54} \right]$$ .........$ Using $(1)$
$ \Rightarrow 210 = \frac{n}{2}(70)$
$ \Rightarrow 210 = 35n$
$ \Rightarrow n = \frac{{210}}{{35}}$
$ \Rightarrow n = 6$
Putting $n = 6$ in equation $(1)$, we get
$(6 - 1)d = 54$
$ \Rightarrow 5d = 54$
$ \Rightarrow d = \frac{{54}}{5}$
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