Algebra
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Q.

A man arranges to pay off a debt of Rs 3600 by 40 annual installments which are in A.P. When 30 of the installments are paid he dies leaving one-third of the debt unpaid.

The value of the $8^{th}$ installment is:

 A.

Rs 35

 B.

Rs 50

 C.

Rs 65

 D.

Rs 70

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Solution:
Option(C) is correct

Let the first installment be '$ a $' and the common difference between any two consecutive installments be '$ d $'

Using the formula for the sum of an A.P.
\[S=\dfrac{n}{2}[2a+(n−1)d]\]
We have, 
\begin{align*}
3600&= \dfrac{40}{2}[2a+(40−1)d]\\
\Rightarrow 180&=2a+39d \tag{1} \label{eq1}\\
\text{And }2400&= \dfrac{30}{2}[2a+(30−1)d] \\
\Rightarrow 160&=2a+29d \tag{2} \label{eq2}
\end{align*}
On solving both the equations we get:
$ d=2 $ and $ a=51 $

Value of $8^{th}$ installment; 

$=51+(8−1)2$ 
$= \text{Rs. } 65$


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