Question

raghav repays a loan of rs. 118000 by paying every month starting with the first installment of rs 1000. he increases the installment with 100 every month. what amount will he pay as the last installmentof loan?
plz answer fast. I want it with steps.

Answer 1

Given that Raghav pays a loan of Rs 118000
First installment Rs1000
And increase Rs 100 for every next month.
With this get an AP: 1000, 1200, 1300
Here 1st term `a` = 1000
Common difference `d` = 100
Total no.of installments or the last month
Total installments will sum up to 118000
∴ Sn = 118000
We know that Sn = n/2[ 2a +(n-1)d]
⇒n/2[ 2a +(n-1)d] = 118000
⇒n [ 2(1000) + (n-1)100] = 118000 ×2
⇒n [ 2000 +100n - 100] = 236000
⇒n [ 1900 + 100n ] = 236000
⇒1900n +100n² = 236000
⇒100n² + 1900n -236000 = 0
⇒100(n² + 19n - 2360) = 0
⇒n² +19n -2360 = 0/100
⇒n² +19n -2360 = 0
⇒n² -40n + 59n - 2360 = 0
⇒n(n -40) +59(n-40) = 0
⇒(n-40) ( n+59) = 0
⇒n-40 = 0                      or           n+59 = 0
⇒n = 40                         or           n = -59
∴n =40 [as n can not be negative here]
Therefore last month he paid = 40th month
term 40  = a+(n-1)d
              = 1000+(40-1)100
              = 1000+39(100)
              = 1000+3900
              = 4000
∴The amount he paid in his last installment is Rs4000