Answer to Question #325660 in Algebra for Azuazu Roy

Question

1. A man saves #100 in his first year of work and each year he saves #20 more than the proceeding year. In how many years will be save #5800.

2. A polygon has 25 sides, the lengths of which starting from the smallest sides are in A.P, if the perimeter of the polygon is 2100cm and the length of the largest side is 20times that of the smallest.find the length of the smallest side and the common difference of the A.P

Answer 1

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What he saved first called first a=100

Common different d=20

Sum of the last term S_n=5800

Using the formular $S_n=\frac{n}{2}(2a+(n-1)d)$

$5800 = \frac{n}{2}(2×100+(n-1)20)$

$5800 = \frac{n}{2}(200+20n-20)\\ 5800 = \frac{n}{2}(180+20n)\\ 5800 = \frac{20×n}{2}(9+n)\\ 580 = n(9+n)\\ n^2 +9n-580 = 0$

In the quadratic equation ax² + bx + c = 0

a = 1; b = 9; c = -580

$D = b^2 -4ac \\ x = \frac{-b±\sqrt{D}}{2a}\\ D = 9^2 - 4×1×(-580) = 81 + 2320 =2401\\ \sqrt{D} = \sqrt{2401} = 49\\ x_1 = \frac{-9-49}{2×1} = -29\\ x_2 = \frac{-9+49}{2×1} = 20$

n₁ = -29 (incorrect)

n₂ = 20

Answer: 20 years

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Polygon has 25 terms which are in AP

Given,

Perimeter of polygon = 2100

Means sum of all terms AP = 2100

Hence,

If t₁, t_n are first and last term AP

Then, given

$t_n = 20t_1\\ 2100 = 25/2(t_1 + t_n)\\ 2100 = 25/2(t_1 + 20t_1)\\ 2100 = 25/2×(21t_1)\\ 100 = 25t_1/2\\ t_1= 200/25 = 8\\ t_{25} = 8 + (25-1)d\\ 20*8 = 8 + 24d\\ 24d = 152\\ d = 19/3\\$

Answer:

the smallest side = 8

the common difference = 19/3

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