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Mekhanik [1.2K]
3 years ago
10

A man has saved $640 during the first month,$720 in the second month and $800 in the third month. If he continues his savings in

this sequence, what will be his savings in the 25th month? *
Mathematics
1 answer:
bija089 [108]3 years ago
8 0

Answer:

Step-by-step explanation:

An arithmetic progression is a list of numbers a1, a2, a3 ………….. an in which each term is obtained by adding a fixed number to the preceding term except the first term.

This fixed number is called the common difference( d ) of the AP. Common difference of an AP will be the difference between any two consecutive terms.

a2= a1+d

a3= a2+d

a4= a3+d

……..

an= an-1+d ………

Each of the numbers in the list is called a term .

Method to find the common difference :

d = a2 - a1 or a3 - a2 or a4 - a3...

General form of an AP.:

a, a+d, a+2d, a+3d…….

Here a is the first term and d is common difference.

General term or nth term of A.P

The general term or nth term of A.P is given by an or tn = a + (n – 1)d, where a = a1 is the first term, d is the common difference and n is the number of term.

SOLUTION :

Given -

A man saved in the first month,in the second month ,in the third month… are 640, 720, 800 .. which forms a sequence(AP).

Here, a1 or t1 = 640 , a2 or t2= 720, a3 or t3 = 800

d = t2 – t1

d= 720- 640

d= 80

tn = a + (n-1) d

t25 = 640 + (25 - 1) 80

t25 = 640 + 24 (80)

t25= 640 + 1920

t25 = 2560

Hence, his Saving will be 2560 in the 25th month.

Hope this helps,from Armax.

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2 years ago
a new car, originally worth $35,795, depreciates at a rate of 17% per year. The value of the car can be represented by the equat
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We have been given that a new car, originally worth $35,795, depreciates at a rate of 17% per year. The value of the car can be represented by the equation y=35795(0.83)^x, where x represents the number of years since purchase and y represents the value (in dollars) of the car.

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3 years ago
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