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Case Study - 1
A woman borrowed ₹10,00,000 from her friend and promised to return the borrowed money in monthly instalments beginning from the next month. After one month, she returned ₹10,000, the next month she returned ₹15,000, the third month she returned ₹20,000 and so on, thereby increasing the monthly instalment uniformly.
Based on the above information, answer the following questions :
(i) Find the amount of instalment paid in the tenth month.
(ii) In which instalment did she pay ₹40,000 ?
(iii) (a) If she returned ₹11,50,000 in all, how many instalments did she pay ?
OR
(b) By which instalment has she returned a total amount of ₹3,25,000 ?
A woman borrowed ₹10,00,000 from her friend and promised to return the borrowed money in monthly instalments beginning from the next month. After one month, she returned ₹10,000, the next month she returned ₹15,000, the third month she returned ₹20,000 and so on, thereby increasing the monthly instalment uniformly.
Based on the above information, answer the following questions :
(i) Find the amount of instalment paid in the tenth month.
(ii) In which instalment did she pay ₹40,000 ?
(iii) (a) If she returned ₹11,50,000 in all, how many instalments did she pay ?
OR
(b) By which instalment has she returned a total amount of ₹3,25,000 ?
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Solution: (i) $a_{10} = 10000 + 5000 \times 9 = 55000$
$\Rightarrow$ Amount of instalment paid in the tenth month $=\text{Rs} 55,000$
(ii) $40000 = 10000 + (n - 1) 5000 \Rightarrow n = 7$
$\Rightarrow$ In the $7^{th}$ instalment she paid ₹40,000
(iii) (a) $1150000 = \frac{n}{2} [20000 + (n - 1) 5000]$
$5n^2 + 15n - 2300 = 0$ or $n^2 + 3n - 460 = 0$
$(n - 20) (n + 23) = 0 \Rightarrow n = 20, n = -23$ (rejected)
$\Rightarrow$ She paid $20$ instalments in order to return ₹11,50,000
OR
(iii) (b) $325000 = \frac{n}{2} [20000 + (n - 1) 5000]$
$5n^2 + 15n - 650 = 0$ or $n^2 + 3n - 130 = 0$
$(n - 10) (n + 13) = 0 \Rightarrow n = 10, n = -13$ (rejected)
$\Rightarrow$ She returned ₹3,25,000 by the $10^{th}$ instalment.
$\Rightarrow$ Amount of instalment paid in the tenth month $=\text{Rs} 55,000$
(ii) $40000 = 10000 + (n - 1) 5000 \Rightarrow n = 7$
$\Rightarrow$ In the $7^{th}$ instalment she paid ₹40,000
(iii) (a) $1150000 = \frac{n}{2} [20000 + (n - 1) 5000]$
$5n^2 + 15n - 2300 = 0$ or $n^2 + 3n - 460 = 0$
$(n - 20) (n + 23) = 0 \Rightarrow n = 20, n = -23$ (rejected)
$\Rightarrow$ She paid $20$ instalments in order to return ₹11,50,000
OR
(iii) (b) $325000 = \frac{n}{2} [20000 + (n - 1) 5000]$
$5n^2 + 15n - 650 = 0$ or $n^2 + 3n - 130 = 0$
$(n - 10) (n + 13) = 0 \Rightarrow n = 10, n = -13$ (rejected)
$\Rightarrow$ She returned ₹3,25,000 by the $10^{th}$ instalment.