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Solve the following:

1. even integers between 1 and 101
2. first 25 terms of the arithmetic sequence 4, 9, 14, 19, 24 ...​

Sagot :

mahesharanthilini

[tex]\huge\mathcal {♨Answer♥}[/tex]

[tex]\large\texttt {1. even integers between }[/tex]

[tex]\large\texttt {1 and 101}[/tex]

The even numbers from 1 to 100 are,

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98 and 100.

[tex]\large\texttt {2. first 25 terms of the }[/tex]

[tex]\large\texttt {arithmetic sequence 4, 9, }[/tex]

[tex]\large\texttt {14, 19, 24 ...}[/tex]

arithmetic sequence = 4, 9, 14, 19, 24 ...

Difference(d) = 9 - 4

= 5

Let's find the first 25 terms of the sequence...

1st term[tex]\large\texttt {↪}[/tex] 4

2nd term[tex]\large\texttt {↪}[/tex] 4 + 5 = 9

3rd term[tex]\large\texttt {↪}[/tex] 9 + 5 = 14

4th term[tex]\large\texttt {↪}[/tex] 14 + 5 = 19

5th term[tex]\large\texttt {↪}[/tex] 19 + 5 =24

6th term[tex]\large\texttt {↪}[/tex] 24 + 5 = 29

7th term[tex]\large\texttt {↪}[/tex] 29 + 5 = 34

8th term[tex]\large\texttt {↪}[/tex] 34 + 5 = 39

9th term[tex]\large\texttt {↪}[/tex] 39 + 5 = 44

10th term[tex]\large\texttt {↪}[/tex] 44 + 5 = 49

11th term[tex]\large\texttt {↪}[/tex] 49 + 5 = 54

12th term[tex]\large\texttt {↪}[/tex] 54 + 5 = 59

13th term[tex]\large\texttt {↪}[/tex] 59 + 5 = 64

14th term[tex]\large\texttt {↪}[/tex] 64 + 5 = 69

15th term[tex]\large\texttt {↪}[/tex] 69 + 5 = 74

16th term[tex]\large\texttt {↪}[/tex] 74 + 5 = 79

17th term[tex]\large\texttt {↪}[/tex] 79 + 5 = 84

18th term[tex]\large\texttt {↪}[/tex] 84 + 5 = 89

19th term[tex]\large\texttt {↪}[/tex] 89 + 5 = 94

20th term[tex]\large\texttt {↪}[/tex] 94 + 5 = 99

21st term[tex]\large\texttt {↪}[/tex] 99 + 5 = 104

22nd term[tex]\large\texttt {↪}[/tex] 104 + 5 = 109

23rd term[tex]\large\texttt {↪}[/tex] 109 + 5 = 114

24th term[tex]\large\texttt {↪}[/tex] 114 + 5 = 119

25th term[tex]\large\texttt {↪}[/tex] 119 + 5 = 124

☆...hope this helps...☆

_♡_mashi_♡_

semsee45

Answer:

1.  2550

2.  1600

Step-by-step explanation:

Question 1

Even Integer: A whole number that can be positive, negative, or zero, and that can be divided exactly by 2.

Even integers between 1 and 101:  

  • 2, 4, 6, 8, 10, ... , 100

This forms an arithmetic sequence with:

  • first term (a) = 2
  • common difference (r) = 2

Sum of the first n terms of an arithmetic series:

[tex]\boxed{S_n=\dfrac12n[2a+(n-1)d]}[/tex]

There are 50 even integers between 1 and 101.  

Therefore, to find the sum of the even integers between 1 and 101, substitute n = 50 and the found values of a and r into the formula:

[tex]\begin{aligned}\implies S_{50} & =\dfrac{1}{2}(50)[2(2)+(50-1)2]\\\\& =25[4+(49)2]\\\\& =25[4+98]\\\\& =25[102]\\\\& = 2550\end{aligned}[/tex]

Therefore, the sum of the even integers between 1 and 100 is 2550.

Question 2

Given arithmetic sequence:

  • 4, 9, 14, 19, 24, ...

From inspection of the given arithmetic sequence:

  • first term (a) = 4
  • common difference (r) = 9 - 4 = 5

Sum of the first n terms of an arithmetic series:

[tex]\boxed{S_n=\dfrac12n[2a+(n-1)d]}[/tex]

To find the sum of the first 25 terms of the given arithmetic sequence, substitute n = 25 and the found values of a and r into the formula:

[tex]\begin{aligned}\implies S_{25} & =\dfrac{1}{2}(25)[2(4)+(25-1)5]\\\\& =\dfrac{25}{2}[8+(24)5]\\\\& =\dfrac{25}{2}[8+120]\\\\& =\dfrac{25}{2}[128]\\\\& = \dfrac{3200}{2}\\\\& = 1600\end{aligned}[/tex]

Therefore, the sum of the first 25 terms of the given arithmetic sequence is 1600.

Learn more about arithmetic series here:

https://brainly.com/question/28350783

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