Question

a man saves rs32 during the first year,rs36 during the second year and this way he increase his savings rs4 per year. find in that time his savings will be rs200?​

Answer 1

Answer:

50 years

Step-by-step explanation:

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Answer 2

HEY USER !

In 5 years a man saving will be ₹ 200.

Step-by-step explanation:

Given :

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

, Sn = 200 Let 'n' be the number of years.

, By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

=>200 = n/2 [2 × 32 + (n– 1)4]

=>200 = n/2 [2 × 32 + (n– 1)4]200 × 2 = n[64 + 4(n - 1)]

=>200 = n/2 [2 × 32 + (n– 1)4]200 × 2 = n[64 + 4(n - 1)]400 = n[64 + 4n - 4]

=>200 = n/2 [2 × 32 + (n– 1)4]200 × 2 = n[64 + 4(n - 1)]400 = n[64 + 4n - 4]400 = n[60 + 4n]

=>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]

=>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

=>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 + n²100 = 15n + n²

=>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 + n²100 = 15n + n²n² + 15n – 100 = 0

=>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 + n²100 = 15n + n²n² + 15n – 100 = 0n² + 20n – 5n – 100 = 0

[By middle term splitting]

n (n + 20) – 5 (n + 20) = 0

n (n + 20) – 5 (n + 20) = 0(n – 5) (n + 20) = 0

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

Number of years can never be negative. Therefore, n = 5Hence, in 5 years a man saving will be ₹ 200.

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