An arithmetic series consists of five terms; the second term is 7 and the last term is 22. write out the full series.
T5 = T2 + 3d, so d = (22-7)/3 = 5
The sequence is
2 7 12 27 22
Technically, the above is an arithmetic sequence. If you truly wanted an arithmetic series, then you'd have to use the sums
S2 = 2/2(2a+d) = 7
S5 = 5/2(2a+4d) = 22
a = 16/5 and d = 3/5 The sequence is
16/5 19/5 22/5 5 28/5
and the series (the sequence of partial sums) is
16/5 7 57/5 82/5 22
To write out the full series, we need to find the common difference (d) between each term.
The formula for the nth term of an arithmetic series is:
\( T_n = a + (n - 1)d \),
where \( T_n \) represents the nth term, \( a \) is the first term, and \( d \) is the common difference.
In this case, we know that the second term is 7, so we can substitute \( T_2 = 7 \) into the formula:
\( 7 = a + (2 - 1)d \).
We also know that the last term is 22, so we can substitute \( T_5 = 22 \) into the formula:
\( 22 = a + (5 - 1)d \).
Simplifying these equations, we get a system of equations:
\( a + d = 7 \) -- Equation 1
\( a + 4d = 22 \) -- Equation 2
Now, let's solve this system of equations to find the values of \( a \) and \( d \).
Subtracting Equation 1 from Equation 2, we get:
\( a + 4d - a - d = 22 - 7 \)
\( 3d = 15 \)
Dividing both sides by 3, we find that \( d = 5 \).
Substituting this value of \( d \) into Equation 1, we have:
\( a + 5 = 7 \)
Simplifying, we find that \( a = 2 \).
Therefore, the first term (\( a \)) is 2 and the common difference (\( d \)) is 5.
Now, let's write out the full series:
\( T_1 = a = 2 \)
\( T_2 = a + d = 2 + 5 = 7 \)
\( T_3 = a + 2d = 2 + 2(5) = 2 + 10 = 12 \)
\( T_4 = a + 3d = 2 + 3(5) = 2 + 15 = 17 \)
\( T_5 = a + 4d = 2 + 4(5) = 2 + 20 = 22 \)
Thus, the full series is: 2, 7, 12, 17, 22.