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Arithmetic Progressions — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsArithmetic Progressions3 Marks
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

$a=2, d=8, S_n=90$
$S_n=\frac{n}{2}[2 a+(n-1) d] \Rightarrow 90=\frac{n}{2}[2 \times 2+(n-1) 8]$
$\Rightarrow 180=n(4+8 n-8) \Rightarrow 180=n(8 n-4)$
$\Rightarrow 8 n^2-4 n-180=0 $
$\Rightarrow 2 n^2-n-45=0 \text { (Dividing by 4) }$
$\Rightarrow 2 n^2-10 n+9 n-45=0$
$\begin{Bmatrix}\because-45\times2=-90\\ \therefore-90=-10\times9\\ -1=-10+9 \end{Bmatrix}$
$\Rightarrow 2 n(n-5)+9(n-5)=0 $
$\Rightarrow(n-5)(2 n+9)=0$
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 n=5 $
$a_n=a+(n-1) d=2+(5-1) \times 8 $
$=2+4 \times 8=2+32=34$

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