Question

In an arithmetic series S5=180=S12,where S5 and S12 are the sums of five and twelve terms respectively calculate fitheeth term of re series

Answer 1

In an arithmetic series, the sum of the first \( n \) terms can be calculated using the formula:

\[
S_n = \frac{n}{2} (2a + (n-1)d)
\]

where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( a \) is the first term,
- \( d \) is the common difference, and
- \( n \) is the number of terms.

Given the information:
\[
S_5 = 180
\]
\[
S_{12} = 180
\]

We can write the equations for the sums:

1. From the first equation for \( n = 5 \):
\[
S_5 = \frac{5}{2} (2a + 4d) = 180
\]

Multiplying through by 2:
\[
5(2a + 4d) = 360
\]
\[
2a + 4d = 72 \quad \text{(1)}
\]

2. From the second equation for \( n = 12 \):
\[
S_{12} = \frac{12}{2} (2a + 11d) = 180
\]

Multiplying through by 2:
\[
12(2a + 11d) = 360
\]
\[
2a + 11d = 30 \quad \text{(2)}
\]

Now, we will simplify the system of equations (1) and (2).

From equation (1):
\[
2a + 4d = 72
\]
From equation (2):
\[
2a + 11d = 30
\]

Now, we can subtract equation (1) from equation (2):
\[
(2a + 11d) - (2a + 4d) = 30 - 72
\]
\[
7d = -42
\]
\[
d = -6
\]

Now, substitute \( d \) back into equation (1) to find \( a \):
\[
2a + 4(-6) = 72
\]
\[
2a - 24 = 72
\]
\[
2a = 96
\]
\[
a = 48
\]

Now we have \( a = 48 \) and \( d = -6 \).

To find the fifteenth term of the series \( a_n \):
\[
a_n = a + (n-1)d
\]
For \( n = 15 \):
\[
a_{15} = 48 + (15 - 1)(-6)
\]
\[
= 48 + 14(-6)
\]
\[
= 48 - 84
\]
\[
= -36
\]

Thus, the fifteenth term of the series is:
\[
\boxed{-36}
\]