A man saved rs 32 during the first year, rs 36 in the second year and in this way he increases his savings by rs 4 every year.find in what time his saving will be rs 200.
Answers
Answer:
In first year a man saved, a = ₹ 32
In second year a man saved , a2 = ₹ 36
Increase in savings, d = 36 - 32 = 4
His saving in 'n' years will be , Sn = 200
Let 'n' be the number of years.
By using the formula ,Sum of n terms,S
n
=(
2
n
)[2a+(n–1)d]
200=n/2[2×32+(n–1)4]
200×2=n[64+4(n−1)]
400=n[64+4n−4]
400=n[60+4n]
400=4n[15+n]
400/4=15n+n²
100=15n+n²
n²+15n–100=0
n²+20n–5n–100=0
[By middle term splitting]
n(n+20)–5(n+20)=0
(n–5)(n+20)=0
(n – 5) = 0, n = 5 & (n + 20) = 0, n = - 20
Number of years can never be negative. Therefore, n = 5
Hence, in 5 years a man saving will be ₹ 200.
Answer:
₹ 200.
Step-by-step explanation:
In first year a man saved, a = ₹ 32
In second year a man saved , a2 = ₹ 36
Increase in savings, d = 36 - 32 = 4
His saving in 'n' years will be , Sn = 200
Let 'n' be the number of years.
By using the formula ,Sum of n terms,S
n
=(
2
n
)[2a+(n–1)d]
200=n/2[2×32+(n–1)4]
200×2=n[64+4(n−1)]
400=n[64+4n−4]
400=n[60+4n]
400=4n[15+n]
400/4=15n+n²
100=15n+n²
n²+15n–100=0
n²+20n–5n–100=0
[By middle term splitting]
n(n+20)–5(n+20)=0
(n–5)(n+20)=0
(n – 5) = 0, n = 5 & (n + 20) = 0, n = - 20
Number of years can never be negative. Therefore, n = 5
Hence, in 5 years a man saving will be ₹ 200.