Question
Let there be an $A.P.$ with first term $'a',$ common difference $'d'.$ If $a_n$ denotes in $n^{th}$ term and $S_n$ the sum of first $n$ terms, find.
$a,$ if $a_n= 28, S_n= 144$ and $n = 9.$
$a,$ if $a_n= 28, S_n= 144$ and $n = 9.$
✓
Answer
$ a_n=28, S_n=144, n=9 $
$ a_n=a+(n-1) d $
$ \Rightarrow a_9=a+(9-1) d=a+8 d$
$a + 8d = 28 .....(i)$
$\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]\Rightarrow\ \frac{9}{2}(\text{a}+28)=144$
$\Rightarrow\ \text{a}+28=\frac{144\times2}{9}=32$
$\text{a}=32-48=4$
$\therefore$ From $(i)$
$4+8\text{d}=28\Rightarrow\ 8\text{d}=28-4=24$
$\Rightarrow\ \text{a}=\frac{24}{8}=3$
Hence $a = 4.$
$ a_n=a+(n-1) d $
$ \Rightarrow a_9=a+(9-1) d=a+8 d$
$a + 8d = 28 .....(i)$
$\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]\Rightarrow\ \frac{9}{2}(\text{a}+28)=144$
$\Rightarrow\ \text{a}+28=\frac{144\times2}{9}=32$
$\text{a}=32-48=4$
$\therefore$ From $(i)$
$4+8\text{d}=28\Rightarrow\ 8\text{d}=28-4=24$
$\Rightarrow\ \text{a}=\frac{24}{8}=3$
Hence $a = 4.$
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