Question
In an AP: $a_3=15, S_{10}=125$, find $d$ and $a_{10}$.
✓
Answer
Here, $a_3=15$
$S_{10}=125$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow a_3 = a + (3 - 1)d$
$ \Rightarrow a_3 = a + 2d$
$ \Rightarrow 15 = a + 2d$
$ \Rightarrow a + 2d = 15 ...... (1)$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{10}} = \frac{{10}}{2}\left[ {2a + (10 - 1)d} \right]$
$ \Rightarrow S_{10} = 5(2a + 9d)$
$ \Rightarrow 125 = 5(2a + 9d)$
$ \Rightarrow 25 = 2a + 9d$
$ \Rightarrow 2a + 9d = 25 ....... (2)$
Solving equation (1) and equation (2), we get
$a = 17$
$d = -1$
Now $an = a + (n - 1)d$
$ \Rightarrow a_{10} = a + (10 - 1)d$
$ \Rightarrow a_{10} = a + 9d$
$ \Rightarrow a_{10} = 17 + 9(-1)$
$ \Rightarrow a_{10} = 17 - 9$
$ \Rightarrow a_{10} = 8$
$S_{10}=125$
We know that
$a_n = a + (n - 1)d$
$ \Rightarrow a_3 = a + (3 - 1)d$
$ \Rightarrow a_3 = a + 2d$
$ \Rightarrow 15 = a + 2d$
$ \Rightarrow a + 2d = 15 ...... (1)$
Again, we know that
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow {S_{10}} = \frac{{10}}{2}\left[ {2a + (10 - 1)d} \right]$
$ \Rightarrow S_{10} = 5(2a + 9d)$
$ \Rightarrow 125 = 5(2a + 9d)$
$ \Rightarrow 25 = 2a + 9d$
$ \Rightarrow 2a + 9d = 25 ....... (2)$
Solving equation (1) and equation (2), we get
$a = 17$
$d = -1$
Now $an = a + (n - 1)d$
$ \Rightarrow a_{10} = a + (10 - 1)d$
$ \Rightarrow a_{10} = a + 9d$
$ \Rightarrow a_{10} = 17 + 9(-1)$
$ \Rightarrow a_{10} = 17 - 9$
$ \Rightarrow a_{10} = 8$
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