Let there be an A.P. with first term ' a ', common difference ' d… - Vidyadip

Question

Let there be an A.P. with first term ' $a$ ', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find.
n and $a _{ n }$, if $a =2, d=8$ and $S _{ n }=90$.

✓

Answer

$a = 2, d = 8, S_n = 90$
$\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\Rightarrow90=\frac{\text{n}}{2}[2\times2+(\text{n}-1)8]$
$\Rightarrow 180 = n(4 + 8n - 8) \Rightarrow 180 = n(8n - 4)$
$\Rightarrow 8n^2 - 4n - 180 = 0$
$\Rightarrow 2n^2 - n - 45 = 0$ (Dividing by $4$)
$\Rightarrow 2n^2 - 10n + 9n - 45 = 0$
$\begin{Bmatrix}\because-45\times2=-90\\ \therefore-90=-10\times9\\ -1=-10+9 \end{Bmatrix}$
$\Rightarrow 2n (n - 5) + 9(n - 5) = 0$
$\Rightarrow (n - 5)(2n + 9) = 0$
Either $n - 5 = 0$, then $n = 5$
or $2n + 9 = 0$,
Then, $2\text{n}=-9\Rightarrow\ \text{n}=\frac{-9}{2}$ but is is not possible being fraction
$\therefore\ \text{n}=5$
$a_n = a + (n - 1)d = 2 + (5 - 1) \times 8$
$= 2 + 4 \times 8 = 2 + 32 = 34$

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