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mars1129 [50]
3 years ago
10

Write a function rule for the following arithmetic sequence and use it to find the 224th term. Show your work.

Mathematics
2 answers:
vagabundo [1.1K]3 years ago
8 0

Answer:

Function rule for the arithmetic sequence is; a_n = a+(n-1)d

The 224th term of the arithmetic sequence  is: 671

Step-by-step explanation:

Arithmetic sequence: A sequence of number which increases or decreases by a constant amount each term.

Formula for nth term of arithmetic sequence is:

a_n = a+(n-1)d

where a is the first term in the sequence, d is the common difference and n is the number of terms;

Given an arithmetic sequence:

2, 5 , 8, 11, ......

First term(a) = 2

Common difference(d) = 3 (Because the difference between the two consecutive terms is 3) i,e

5-2 = 3

8-5 = 3

11 -8 =3 ans so on,....

To find the 224th term;

we have,

a = 2 , d = 3 and n = 224

Using formula for nth arithmetic sequence;

a_{224} = 2+(224-1)(3)

a_{224} = 2+(223)(3)

a_{224} = 2+669 =671

therefore, the 224th term for the given arithmetic sequence is, 671


zavuch27 [327]3 years ago
5 0

Answer:

671  is the 224th term of the given series

Step-by-step explanation:

The given sequence is

2,5,8,11....

First term that is 2

common difference that is 3

we have to find 224th term

That is n =224

using a_n=a+(n-1)d

a_{224}=2+(224-1)3

On solving the above equation we get

a_{224}=671

Hence, 224th term of the given series is 671


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Answer:

(1,41), (2,40), (3,39), (4,38), (5,37), (6,36), (7,35), (8,34), (9,33), (10,32), (11,31), (12,30), (13,29), (14,28), (15,27), (16,26), (17,25), (18,24), (19,23), (20,22), (21,21), (22,20), (23,19), (24,18), (25,17), (26,16), (27,15), (28,14), (29,13), (30,12), (31,11), (32,10), (33,9), (34,8), (35,7), (36,6), (37,5), (38,4), (39,3), (40,2), (41,1).

Step-by-step explanation:

To find all pairs of natural numbers which are solution to a+b=42, we need to first choose a value to one of them (let's choose 'a'), and that value will be the lowest possible, so we begin with a=1, and then we increase to 2, and 3, and so on.

For each value of a, we calculate the value of b using the equation.

So, starting with a=1, we have that b=41

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Doing this until a=41 (so b=1 will be the lowest value possible for b), we have the pairs of natural number we want:

(1,41), (2,40), (3,39), (4,38), (5,37), (6,36), (7,35), (8,34), (9,33), (10,32), (11,31), (12,30), (13,29), (14,28), (15,27), (16,26), (17,25), (18,24), (19,23), (20,22), (21,21), (22,20), (23,19), (24,18), (25,17), (26,16), (27,15), (28,14), (29,13), (30,12), (31,11), (32,10), (33,9), (34,8), (35,7), (36,6), (37,5), (38,4), (39,3), (40,2), (41,1).

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