Question
In an AP: $a = 5, d = 3, a_n = 50$, find n and $S_n.$
✓
Answer
Here, $a = 5, d = 3, a_n = 50$
We know that
$a_n = a + (n – 1)d$
$ \Rightarrow 50 = 5 + (n - 1)3$
$ \Rightarrow (n – 1)3 = 50 - 5$
$ \Rightarrow (n - 1)3 = 45$
$ \Rightarrow n - 1 = \frac{{45}}{3}$
$ \Rightarrow n - 1 = 15$
$ \Rightarrow n = 15 + 1$
$ \Rightarrow n = 16$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_n} = \frac{{16}}{2}\left[ {2(5) + (16 - 1)3} \right]$
$ \Rightarrow S_n = 8[10 +45]$
$ \Rightarrow S_n = 8(55)$
$ \Rightarrow S_n = 440$
We know that
$a_n = a + (n – 1)d$
$ \Rightarrow 50 = 5 + (n - 1)3$
$ \Rightarrow (n – 1)3 = 50 - 5$
$ \Rightarrow (n - 1)3 = 45$
$ \Rightarrow n - 1 = \frac{{45}}{3}$
$ \Rightarrow n - 1 = 15$
$ \Rightarrow n = 15 + 1$
$ \Rightarrow n = 16$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_n} = \frac{{16}}{2}\left[ {2(5) + (16 - 1)3} \right]$
$ \Rightarrow S_n = 8[10 +45]$
$ \Rightarrow S_n = 8(55)$
$ \Rightarrow S_n = 440$
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