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

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Last updated date: 02nd Mar 2024
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IVSAT 2024
Answer
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Hint-Use sum of n terms of an A.P. As the given conditions form an A.P.

According to the question, the total amount of debt to be paid in \[40\] installments is $3600$ pounds
So, the sum of all $40$ installments ${S_{40}} = 3600$
After $30$installments one-third of his debt is left unpaid this means that he paid two third of the payment.
So, the amount he paid in $30$ installments
  $
   = \frac{2}{3} \times 3600 = 2400 \\
    \\
$
Sum of first $30$ installments is ${S_{30}} = 2400$
 Let us take the first installments as $a$ and common difference as $d$.
So, using the formula for the sum of $n$ terms of an A.P.
${S_n} = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Let us find $a$ and$d$ for $30$ installments
$
  {S_{30}} = \frac{{30}}{2}\left[ {2a + \left( {30 - 1} \right)d} \right] = 2400 \\
   \Rightarrow 15\left( {2a + 29d} \right) = 2400 \\
   \Rightarrow 2a + 29d = 160...........\left( 1 \right) \\
$
As we know the sum of all $40$ installments is ${S_{40}} = 3600$
$
  {S_{40}} = \frac{{40}}{2}\left[ {2a + \left( {40 - 1} \right)d} \right] = 3600 \\
   \Rightarrow 20\left( {2a + 39d} \right) = 3600 \\
   \Rightarrow 2a + 39d = 180...........\left( 2 \right) \\
$
Solve for (1) and (2) equations
$
   \Rightarrow 2a + 30d - 2a - 29d = 180 - 160 \\
   \Rightarrow 10d = 20 \\
   \Rightarrow d = 2 \\
 $
Put value of $d$ in any equations
$
   \Rightarrow 2a + 29 \times 2 = 160 \\
   \Rightarrow 2a + 58 = 160 \\
   \Rightarrow 2a = 102 \\
   \Rightarrow a = 51 \\
$
So, the first installment is 51 pounds.

Note- Whenever we come to these types of problems first of all find the sum of all installments and also find the sum of rest of installments then after using sum of $n$ terms of an AP So, we can easily calculate the first installment.
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