Find 8 terms of the arithmetic progression if the second term is 2 and the sixth term is 5

I assume you mean the first 8 terms...

a+d = 2
a+5d = 5

4d = 3
d = 3/4
a = 5/4

5/4 8/4 11/4 14/4 17/4 20/4 23/4 26/4 ...
or,
5/4 2 11/4 7/2 17/4 5 23/4 13/2

it's a number i'm sure of it. i caculated it and the answer is 34

To find the terms of an arithmetic progression, we need to use the formula for the general term:

\[a_n = a_1 + (n-1)d\]

where \(a_n\) is the \(n\)th term of the progression, \(a_1\) is the first term, \(n\) is the position of the term, and \(d\) is the common difference between the terms.

Given that the second term \(a_2\) is 2 and the sixth term \(a_6\) is 5, we can set up two equations:

\[a_2 = a_1 + (2-1)d = a_1 + d = 2\]

\[a_6 = a_1 + (6-1)d = a_1 + 5d = 5\]

Solving these equations simultaneously will allow us to find the values of \(a_1\) and \(d\).

Subtracting the first equation from the second equation gives:

\[(a_1 + 5d) - (a_1 + d) = 5 - 2\]
\[4d = 3\]
\[d = \frac{3}{4}\]

Substituting this value of \(d\) into the first equation:

\(a_1 + \frac{3}{4} = 2\)

Solving for \(a_1\):

\(a_1 = 2 - \frac{3}{4} = \frac{5}{4}\)

Now, we can plug these values back into the general term formula to find the first eight terms of the arithmetic progression:

\(a_n = \frac{5}{4} + (n-1)\left(\frac{3}{4}\right)\)

Substituting \(n = 1\) through 8, we get:

\(a_1 = \frac{5}{4}\)
\(a_2 = \frac{5}{4} + (2-1)\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{3}{4} = 2\)
\(a_3 = \frac{5}{4} + (3-1)\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{6}{4} = \frac{11}{4}\)
\(a_4 = \frac{5}{4} + (4-1)\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{9}{4} = \frac{14}{4} = \frac{7}{2}\)
\(a_5 = \frac{5}{4} + (5-1)\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{12}{4} = \frac{17}{4}\)
\(a_6 = \frac{5}{4} + (6-1)\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{15}{4} = \frac{20}{4} = 5\)
\(a_7 = \frac{5}{4} + (7-1)\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{18}{4} = \frac{23}{4}\)
\(a_8 = \frac{5}{4} + (8-1)\left(\frac{3}{4}\right) = \frac{5}{4} + \frac{21}{4} = \frac{26}{4} = \frac{13}{2}\)

Therefore, the first eight terms of the arithmetic progression are:

\(\frac{5}{4}, 2, \frac{11}{4}, \frac{7}{2}, \frac{17}{4}, 5, \frac{23}{4}, \frac{13}{2}\)