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Arithmetic Progressions — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsArithmetic Progressions4 Marks
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.$n$ and $a$, if $a_n = 4, d = 2$ and $S_n = -14.$
✓

Answer

$a_n=4, d=2, S_n=-14$
$\therefore a+(n-1) 2=4 \Rightarrow a+(n-1)^2=4$
$\Rightarrow a+2 n-2=4 \Rightarrow a+2 n=4+2$
$a+2 n=6 \ldots . . \text { (i) }$
$S_n=\frac{n}{2}[a+1] \Rightarrow-14=\frac{n}{2}[a+4]$
$n(a+4)=-28 \text {.....(ii) }$
From (i) $a=6-2 n$
$\therefore \text { in (ii) }$
$n(6-2 n+4)=-28 \Rightarrow n(10-2 n)=-28$
$\Rightarrow 10 n-2 n^2=-28 \Rightarrow 2 n^2-10 n-28=0$
$\left.\Rightarrow n^2-5 n-14=0 \text { (Divisible by } 2\right)$
$\Rightarrow n^2-7 n+2 n-14=0$
$\begin{Bmatrix} \therefore\ -14=-7\times2\\ -5=-7=2 \end{Bmatrix}$
$\Rightarrow n(n - 7) + 2(n - 7) = 0$
$\Rightarrow (n - 7)(n + 2) = 0$
Either $n-7=0$, then $n=7$
of $n +2=0$, then $n =-2$ but it is not possible being negative
$\therefore n =7$
Now $a =6-2 n =6-2 \times 7=6-14=-8$
Hence $a=-8, n=7$

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