A man repays a loan of Rs. 4290 by paying Rs. 300 in the first month and then decrease the payment by Rs.10 every month. How long it will take to clear his loan?
Answers
Answer:
Given : Sn = 4290, a = 300. d = -10
Sn = n/2[2a+(n-1)d]
4290 = n/2[2(300)+(n-1)-10]
4290 = n/2[600-10n+10]
4290 = n/2[610-10n]
4290 = n/2[10(61-n)]
(4290×2)/10 = n(61-n)
858 = -n²+61n
n²-61n+858 = 0
n²-39n-22n+858 = 0
n(n-39)-22(n-39) = 0
n-39 = 0. or n-22 = 0
n = 39 or n = 22
Question :--- A man repays a loan of Rs. 4290 by paying Rs. 300 in the first month and then decrease the payment by Rs.10 every month. How long it will take to clear his loan ?
Concept and formula used :---
Clearly we can see that, its a Question of AP.
where , 4290 is total sum of AP .
He pay Rs.300 in first month that means this is First term.
Now, he Decrease the payment by Rs.10 , that Means this is Common Difference .
→ Sum of n terms of AP = n/2 [ 2a + (n-1)d]
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Solution :---
→ First Term = a = 300
→ Common Difference = d = (-10)
→ Sum of AP = 4290 .
→ n = ?
Putting values in above Told Formula we get,
→ 4290 = n/2 [ 2*300 + (n-1)(-10) ]
→ 4290 = n/2 [ 600 -10 n + 10 ]
→ 4290 = n/2 [ 610 - 10n ]
→ 8580 = 610n - 10n²
→ 10n² - 610n + 8580 = 0
→ 10(n² - 61n + 858) = 0
→ n² - 61n + 858 = 0
Splitting the Middle term now,
→ n² - 39n -22n + 858 = 0
→ n(n-39) - 22(n-39) = 0
→ (n-39)(n-22) = 0
Putting both Equal to zero now, we get,
→ n-39 = 0. or , n -22 = 0
→ n = 39. n = 22 .
Hence, He can pay loan in both 39 as well as 22 months.