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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 â€˜aâ€™ 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 â€“ 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 â€˜aâ€™ is the first term and â€˜dâ€™ 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) â€“ (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 â€˜aâ€™ and this must be taken care of.

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 â€“ 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 â€˜aâ€™ is the first term and â€˜dâ€™ 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) â€“ (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 â€˜aâ€™ and this must be taken care of.

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