# 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.

Answer

Verified

382.8k+ views

**Hint:**First of all, use the formula of sum of n terms of an A.P, which is \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] to find the sum of 40 installments by substituting n = 40 and \[{{S}_{n}}=3600\]. Then find the sum of the last 10 installments by using \[{{S}_{40}}-{{S}_{30}}=1200\]. By solving these two equations, get the value of ‘a’ that is the first installment.

**Complete step-by-step answer:**

We are given that a man arranges to pay a debt of Rs.3600 by paying 40 annual installations which are in arithmetic progression. It is also given that he dies after paying 30 installments leaving one – third of the total debt unpaid. We have to find the value of the first installment.

Let us consider the first installment paid by the man to be Rs. \[a\].

Since we know that the installments are in A.P. Therefore, let us assume that next installments are as follows:

\[a,a+d,a+2d,a+3d,a+4d......\]

where ‘a’ is the first term and ‘d’ is the common difference of A.P.

We know that the general term of A.P is \[{{a}_{n}}=a+\left( n-1 \right)d\].

Since we know that there are total of 40 installments, therefore,

We get 40th installment as

\[{{a}_{40}}=a+\left( 40-1 \right)d\]

\[=a+39d\]

Hence, we get the 1st to 40th installments as follows

\[a,a+d,a+2d,a+3d.......a+39d\]

We know that the sum of n terms of A.P is \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\].

We are given that the sum of these 40 installments is Rs.3600.

Hence, we get,

\[{{S}_{40}}=\dfrac{40}{2}\left[ 2a+\left( 40-1 \right)d \right]=3600\]

By simplifying the above equation, we get,

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

By dividing 20 on both sides, we get,

\[2a+39d=180.....\left( i \right)\]

Now, we are given that he died after paying the first 30 installments leaving one-third of the debt unpaid. That means the sum of the last 10 installments is equal to one-third of debt. We can write it as

Sum of last 10 installments \[=\dfrac{1}{3}[\text{Total Debt}]\]

Since we know that the total debt is Rs.3600. By substituting its value in the above equation, we get,

Sum of last 10 installments \[=\dfrac{3600}{3}=1200.....\left( ii \right)\]

Now, the sum of last 10 installments = (Sum of total 40 installments) – (Sum of first 30 installments)

\[={{S}_{40}}-{{S}_{30}}\]

We know that the sum of the total of 40 installments \[={{S}_{40}}=Rs.3600\]

And the sum of the first 30 installments \[={{S}_{30}}=\dfrac{30}{2}\left( 2a+\left( 30-1 \right)d \right)\]

We get, \[{{S}_{30}}=15\left( 2a+29d \right)\]

By substituting the value of \[{{S}_{40}}\] and \[{{S}_{30}}\], we get,

Sum of last 10 installments \[=3600-15\left( 2a+29d \right)\]

By substituting the value of the sum of the last 10 installments in equation (ii), we get

\[3600-15\left( 2a+29d \right)=1200\]

Or, \[15\left( 2a+29d \right)=3600-1200=2400\]

By dividing 15 on both sides, we get,

\[\left( 2a+29d \right)=160....\left( iii \right)\]

By substituting equation (iii) from (ii), we get,

\[\left( 2a+39d \right)-\left( 2a+29d \right)=180-160\]

We get, \[\left( 39d-29d \right)=20\]

Or, \[10d=20\]

By dividing 10 on both sides, we get,

\[d=2\]

By substituting the value of d = 2 in equation (i), we get,

\[2a+39\times 2=180\]

Or, \[2a+78=180\]

\[\Rightarrow 2a=180-78\]

\[\Rightarrow 2a=102\]

By dividing 2 on both sides, we get,

\[a=51\]

**As we have assumed that the first installment is Rs a, therefore we get the value of the first installment as Rs. 51.**

**Note:**Some students make this mistake of finding the sum of the last 10 installments by using the formula of \[{{S}_{10}}\] that is \[\dfrac{10}{2}\left( 2a+\left( 10-1 \right)d \right)\] but this is wrong. They must note that \[{{S}_{10}}\] is the formula to find the sum of the first 10 installments and sum of the last 10 installments i.e. \[\left( {{S}_{40}}-{{S}_{30}} \right)\]. In general, \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] is the formula of sum of n terms of A.P starting from first term that is ‘a’ and this must be taken care of.

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which one of the following places is unlikely to be class 8 physics CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Elucidate the structure of fructose class 12 chemistry CBSE

What is pollution? How many types of pollution? Define it