a man saves rupees 320 during the first month rupees 360 in the second month and 400 in 3rd month if he continue his savings in this sequence in how many month will it save rs 20000
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Answered by
10
492 months is the answer
Answered by
20
A.P. = 320, 360, 400...
a = 320
d = 360 - 320 = 40
Sn = 20000
Formula =>
Sn = n/2(2a + (n -1)d)
20000 = n/2(2×320 + (n - 1)40)
20000 = n(320 + (n - 1)20)
20000 = n(320 + 20n - 20)
20000 = 300n + 20n²
20n² + 300n - 20000 = 0
20(n² + 15n - 1000) = 0
n² + 15n - 1000 = 0
n² + 40n - 25n - 1000 = 0
n(n + 40) - 25(n + 40) = 0
(n + 40)(n - 25) = 0
n = -40 or 25
Months can't be negative so it should be positive.
Therefore in 25 months it will save ₹20000
a = 320
d = 360 - 320 = 40
Sn = 20000
Formula =>
Sn = n/2(2a + (n -1)d)
20000 = n/2(2×320 + (n - 1)40)
20000 = n(320 + (n - 1)20)
20000 = n(320 + 20n - 20)
20000 = 300n + 20n²
20n² + 300n - 20000 = 0
20(n² + 15n - 1000) = 0
n² + 15n - 1000 = 0
n² + 40n - 25n - 1000 = 0
n(n + 40) - 25(n + 40) = 0
(n + 40)(n - 25) = 0
n = -40 or 25
Months can't be negative so it should be positive.
Therefore in 25 months it will save ₹20000
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