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 $

Note: To check out the calculations, check comment by Anita.

Value of $8^{th}$ installment; 

$=51+(8−1)2$ 
$= \text{Rs. } 65$


(5) Comment(s)


Priyanka
 ()

How $a = 51$ ,came???


Anita
 ()

It came by solving the two equations arrived at the solution,

$180=2a+39d $ -------- (1)

$160=2a+29d $ -------- (2)

Now, $(1)*29-(2)*39$

$180*29=2a*29+39d*29 $

-$(160*39=2a*39+29d*39)$

$=>180*29- 160*39=2a*29-2a*39$

$=> -1020=-20a$

$=> a=51$

It's not that difficult if you give it a proper try.

Anita
 ()

Although it would be easier to find $d$ first and then find the value of $a$.

To find $d$, just do $(2)-(1)$,

$=>(180-160)=2a+39d -(2a+29d)$

$=> 20=10d$

$=>d=2$

Now, put the value of $d$ in $(1)$,

$180=2a+39*2$

$a=\dfrac{180-78}{2}$

$=\dfrac{102}{2}$

$=51$


Priyanka
 ()

2400 = 30/2 [2a + ( 30 - 1 )d ]

how 2400 came??

please explain


Anita
 ()

Since the man dies after paying 30 instalments and leaving one-third of the debt unpaid.

Total debt to be paid = 3600

Debt remaining= $\frac{1}{3}$ of 3600

$=1200$

So, debt paid in the 30 installments,

$=3600-1200$

$=2400$

Which takes to the below equation,

$2400= \dfrac{30}{2}[2a+(30−1)d]$

Hope this helps.