Question

A man arranges to pay off a debt of RS. $3600$ by $40$ annual installments which form an arithmetic series. When $30$ of the installments are paid, he dies leaving one-third of the debt unpaid, find the value of the first installment.

• A. Rs. $61$
• B. Rs. $41$
• C. Rs. $51$
• D. Rs. $11$

Answer 1

Answer: (C)

According to the question,

Tthe total amount of debit to be paid in 40 installment = Rs 3600

After 30 installment one-third of his debit is left unpaid. This means that he paid two third of the payment. So,

The amount he paid in 30 installments = $\frac{2}{3}\left(3600\right)$

= 2(1200)

= 2400

Let us take the first installment as $a$ and common difference as $d$.

So, using the formula for the sum of $n$ terms of an A.p,

$s_n=\frac{n}{2}\left[2a+(n-1)d\right]$

Let us find $a$ and $d$, for 30 installments.

$s_{30}=\frac{30}{2}\left[2a+(30-1)d\right]$

$2400=15\left[2a+\left(29\right)d\right]$

$\frac{2400}{15}=2a+29d$

$160=2a+29d$

$a=\frac{160-29d}{2}$ .....(1)

Similarly, we find $a$ and $d$ for 40 installment.

$s_{40}=\frac{40}{2}\left[2a+(40-1)d\right]$

$3600=20(2a+(39\left)d\right)$

$\frac{3600}{20}=2a+39d$

$a=\frac{180-39d}{2}$

subtracting (1) from (2), we get

$a-a=\left(\frac{180-39d}{2}\right)-\left(\frac{160-29d}{2}\right)$

$0=\frac{180-39d-160+29d}{2}$ .....(2)

$0=20-10d$

Further solving for $d$

10$d$ = 20

$d=\frac{20}{10}$

$d$ = 2 the value

subtracting the value of $d$ in (1), we get,

$a=\frac{160-29\left(2\right)}{2}$

$=\frac{160-58}{2}$

$=\frac{102}{2}$

= Rs 51