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Hint- Here, we will be using the formulas for last term of an AP and sum of first terms of an AP series.

Given, raghav buys a shop for Rs 1,20,000 and he pays half of the amount in cash in the starting itself and is left with the other half of it which he will pay through annual instalments.

Balance amount to be paid$ = \dfrac{{120000}}{2} = {\text{Rs }}60000$

Also, given the amount of each annual instalment is Rs 5,000

Total number of instalments$ = 12$ and rate of interest is 12 percent

Amount to be paid for first instalment$ = 5000 + $12 percent of the balance amount to be paid$ = 5000 + \dfrac{{12}}{{100}} \times 60000 = 5000 + 7200 = {\text{Rs }}12200$

Now, balance amount to be paid$ = 60000 - 5000 = {\text{Rs }}55000$

Amount to be paid for second instalment$ = 5000 + $12 percent of balance amount to be paid$ = 5000 + \dfrac{{12}}{{100}} \times 55000 = 5000 + 6600 = {\text{Rs }}11600$

Now, balance amount to be paid$ = 55000 - 5000 = {\text{Rs }}50000$

Amount to be paid for third instalment$ = 5000 + $12 percent of balance amount to be paid$ = 5000 + \dfrac{{12}}{{100}} \times 50000 = 5000 + 6000 = {\text{Rs }}11000$

Now, if we observe carefully we will get to know that the total amount paid for instalments annually forms arithmetic progression (AP) series because the balance amount to be paid forms an AP. Therefore, the AP series formed by the amount paid for instalments is 12200, 11600, 11000, â€¦ up to 12 instalments.

For this AP series, first term ${a_1} = 12200$, common difference $d = 11600 - 12200 = - 600$and total number of terms$ = 12$

Therefore, the total amount paid for these 12 annual instalments can be obtained by finding the sum of all the terms in the AP series which is given by ${{\text{S}}_n} = \dfrac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$

Total amount paid for these 12 annual instalments$ = \dfrac{{12}}{2}\left[ {2 \times 12200 + \left( {12 - 1} \right)\left( { - 600} \right)} \right] = 6\left( {24400 - 6600} \right) = {\text{ Rs 106800}}$

Total cost of the shop is equal to the sum of the amount paid initially in cash (Rs 60,000) and the total amount paid for these 12 annual instalments (Rs 1,06,800).

i.e., Total cost of the shop$ = 60000 + {\text{106800}} = {\text{Rs 166,800}}$

Therefore, the actual price which raghav has to pay in order to own the shop is Rs 1,66,800.

Note- In these type of problems in which instalments are there, the total price at which the shop was bought is more than the one quoted for that particular shop because of the interest amount that has to be paid by the buyer at the quoted rate of interest according to the time requested for the completion of instalments.

Given, raghav buys a shop for Rs 1,20,000 and he pays half of the amount in cash in the starting itself and is left with the other half of it which he will pay through annual instalments.

Balance amount to be paid$ = \dfrac{{120000}}{2} = {\text{Rs }}60000$

Also, given the amount of each annual instalment is Rs 5,000

Total number of instalments$ = 12$ and rate of interest is 12 percent

Amount to be paid for first instalment$ = 5000 + $12 percent of the balance amount to be paid$ = 5000 + \dfrac{{12}}{{100}} \times 60000 = 5000 + 7200 = {\text{Rs }}12200$

Now, balance amount to be paid$ = 60000 - 5000 = {\text{Rs }}55000$

Amount to be paid for second instalment$ = 5000 + $12 percent of balance amount to be paid$ = 5000 + \dfrac{{12}}{{100}} \times 55000 = 5000 + 6600 = {\text{Rs }}11600$

Now, balance amount to be paid$ = 55000 - 5000 = {\text{Rs }}50000$

Amount to be paid for third instalment$ = 5000 + $12 percent of balance amount to be paid$ = 5000 + \dfrac{{12}}{{100}} \times 50000 = 5000 + 6000 = {\text{Rs }}11000$

Now, if we observe carefully we will get to know that the total amount paid for instalments annually forms arithmetic progression (AP) series because the balance amount to be paid forms an AP. Therefore, the AP series formed by the amount paid for instalments is 12200, 11600, 11000, â€¦ up to 12 instalments.

For this AP series, first term ${a_1} = 12200$, common difference $d = 11600 - 12200 = - 600$and total number of terms$ = 12$

Therefore, the total amount paid for these 12 annual instalments can be obtained by finding the sum of all the terms in the AP series which is given by ${{\text{S}}_n} = \dfrac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$

Total amount paid for these 12 annual instalments$ = \dfrac{{12}}{2}\left[ {2 \times 12200 + \left( {12 - 1} \right)\left( { - 600} \right)} \right] = 6\left( {24400 - 6600} \right) = {\text{ Rs 106800}}$

Total cost of the shop is equal to the sum of the amount paid initially in cash (Rs 60,000) and the total amount paid for these 12 annual instalments (Rs 1,06,800).

i.e., Total cost of the shop$ = 60000 + {\text{106800}} = {\text{Rs 166,800}}$

Therefore, the actual price which raghav has to pay in order to own the shop is Rs 1,66,800.

Note- In these type of problems in which instalments are there, the total price at which the shop was bought is more than the one quoted for that particular shop because of the interest amount that has to be paid by the buyer at the quoted rate of interest according to the time requested for the completion of instalments.

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