Question 921977
General term of sequence, {{{305-(n-1)15}}}.  The last allowable term must be {{{305=15(n-1)}}}
{{{305=15n-15}}}
{{{320=15n}}}
{{{n=21}}} for practical purposes.  Cannot allow for more than 21 monthly payments.


You want to know for what sum of consecutive terms the loan of Rs 3250 will be paid.


{{{(n/2)(305+(305-15(n-1)))}}} compared to 3250;
And remember {{{n<=21}}}.


{{{(n/2)(610-15(n-1))=3250}}}
the steps will lead to
{{{3n^2-125n+1300=0}}}
and the discriminant is 25.


Directly using general solution of a quadratic equation,
{{{n=(125+- sqrt(25))/(2*3)}}}
simplifying to either {{{highlight(n=20)}}}  or {{{n=21&2/3}}}.


Only one of those makes the right kind of sense.  Remember, n must not be more than 21.  

n=20, meaning twenty monthly payments will finish the debt.