Loan 5

Abdul takes a loan of 200000 from Ali and agrees to repay in number of instalment, each instalment begin with the 2nd exceeding the previous one by 1000, if the first instalment is 500, find how many instalment will be necessary to be wipe out
the loan? Completely

Result

n =  20

Solution:

a=200000 a1=500 d=1000  a=a1+an2n a=a1+a1+(n−1)⋅ d2⋅ n  200000∗2=(500∗2+(n−1)∗1000)∗n  200000⋅ 2=(500⋅ 2+(n−1)⋅ 1000)⋅ n −1000n2+400000=0 1000n2−400000=0  a=1000;b=0;c=−400000 D=b2−4ac=02−4⋅1000⋅(−400000)=1600000000 D>0  n1,2=−b±D2a=±16000000002000 n1,2=±400002000 n1,2=±20 n1=20 n2=−20   Factored form of the equation:  1000(n−20)(n+20)=0  n=n1=20a=200000 \ \\ a_{1}=500 \ \\ d=1000 \ \\ \ \\ a=\dfrac{ a_{1}+a_n }{ 2 } n \ \\ a=\dfrac{ a_{1}+a_{1}+(n-1) \cdot \ d }{ 2 } \cdot \ n \ \\ \ \\ 200000 * 2=(500*2 + (n-1)*1000)*n \ \\ \ \\ 200000 \cdot \ 2=(500 \cdot \ 2 + (n-1) \cdot \ 1000) \cdot \ n \ \\ -1000n^2 +400000=0 \ \\ 1000n^2 -400000=0 \ \\ \ \\ a=1000; b=0; c=-400000 \ \\ D=b^2 - 4ac=0^2 - 4\cdot 1000 \cdot (-400000)=1600000000 \ \\ D>0 \ \\ \ \\ n_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ \pm \sqrt{ 1600000000 } }{ 2000 } \ \\ n_{1,2}=\dfrac{ \pm 40000 }{ 2000 } \ \\ n_{1,2}=\pm 20 \ \\ n_{1}=20 \ \\ n_{2}=-20 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 1000 (n -20) (n +20)=0 \ \\ \ \\ n=n_{1}=20

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