Answers

2015-12-19T07:08:21+05:30
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
1 5 1
please understand well and mark as the best answer
its actually 4900
YEAH A MISTAKE