A person has to clear a debt of Rs. 3250 by m | vedclass

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(D) The amounts paid in monthly instalments form an Arithmetic Progression ($A$.$P$.).First term $a = 20$.Common difference $d = 15$.Total debt $S_n =

Vedclass · Accepted Answer

$20$

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(D) The amounts paid in monthly instalments form an Arithmetic Progression ($A$.$P$.).First term $a = 20$.Common difference $d = 15$.Total debt $S_n = 3250$.The sum of $n$ terms of an $A$.$P$. is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Substituting the values: $3250 = \frac{n}{2}[2(20) + (n-1)15]$.$6500 = n[40 + 15n - 15]$.$6500 = n[25 + 15n]$.$15n^2 + 25n - 6500 = 0$.Dividing by $5$: $3n^2 + 5n - 1300 = 0$.Factoring the quadratic equation: $3n^2 + 65n - 60n - 1300 = 0$.$n(3n + 65) - 20(3n + 65) = 0$.$(n - 20)(3n + 65) = 0$.Since $n$ must be a positive integer,$n = 20$.Thus,it will take $20$ months to clear the debt.

$A$ person has to clear a debt of $Rs. 3250$ by monthly instalments. In the first month he has to pay $Rs. 20$ and then onwards every instalment is $Rs. 15$ more than the previous instalment. How long will it take to clear the debt?

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