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