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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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If the sum of certain number of terms of the A.P., $25,22,19................$ is $116$, then find the last term.

$\begin{array}{1 1} 7 \\ 1 \\ 4 \\ -2 \end{array} $

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Toolbox:
  • Sum of $n$ terms of an $A.P.=S_n=\large\frac{n}{2}$$[(2a+(n-1)d]$
  • $n^{th}\:term=t_n=a+(n-1)d$
Given that sum of $n$ terms of the $A.P.$ is $116$
$i.e.,\:\:S_n= 25+22+19+........+t_n=116$
In this A.P. first term $a=25,$ common difference $d=22-25=-3$
We know that $S_n=\large\frac{n}{2}$$[(2a+(n-1)d]$
$\therefore\:116=\large\frac{n}{2}$$[(2\times 25+(n-1)(-3)]$
$\Rightarrow\:232=n(-3n+53)$
$\Rightarrow\:3n^2-53n+232=0$
$\Rightarrow\:3n^2-24n-29n+232=0$
$\Rightarrow\:n=8\:\:or\:\:n=\large\frac{29}{3}$
But since the number of terms $n$ cannot be fraction, $n=8$
$\therefore \:$ The last term $t_n=a+(n-1)d$
$=25+(8-1)(-3)=4$
answered Feb 22, 2014 by rvidyagovindarajan_1
 

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