Question

The sum of 8th term of an arithmetic progression is 160 why the sum of 20th term is 880 find the 43rd Term
the sum of 12th term

Answer 1

To solve the problem, let's denote the first term of the arithmetic progression (AP) as \( a \) and the common difference as \( d \).

The formula for the \( n \)-th term of an AP is given by:

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

The sum of the first \( n \) terms of an arithmetic progression is given by:

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

We are provided the following conditions:

1. The sum of the first 8 terms is 160:

\[
S_8 = \frac{8}{2} (2a + 7d) = 4(2a + 7d) = 160
\]

This simplifies to:

\[
2a + 7d = 40 \quad \text{(1)}
\]

2. The sum of the first 20 terms is 880:

\[
S_{20} = \frac{20}{2} (2a + 19d) = 10(2a + 19d) = 880
\]

This simplifies to:

\[
2a + 19d = 88 \quad \text{(2)}
\]

Now we have a system of two equations:

1. \( 2a + 7d = 40 \)
2. \( 2a + 19d = 88 \)

We can eliminate \( 2a \) by subtracting equation (1) from equation (2):

\[
(2a + 19d) - (2a + 7d) = 88 - 40
\]

This simplifies to:

\[
12d = 48
\]

Solving for \( d \):

\[
d = 4
\]

Now, substituting \( d = 4 \) back into equation (1):

\[
2a + 7(4) = 40
\]

\[
2a + 28 = 40
\]

\[
2a = 12
\]

\[
a = 6
\]

Now we have our values for \( a \) and \( d \):
- \( a = 6 \)
- \( d = 4 \)

Next, we can find the 43rd term, \( a_{43} \):

\[
a_{43} = a + (43-1)d = 6 + 42 \times 4
\]

Calculating it:

\[
a_{43} = 6 + 168 = 174
\]

Finally, we can find the sum of the first 12 terms, \( S_{12} \):

\[
S_{12} = \frac{12}{2} (2a + 11d) = 6(2a + 11d)
\]

Substituting in our values for \( a \) and \( d \):

\[
S_{12} = 6(2(6) + 11(4)) = 6(12 + 44) = 6 \times 56 = 336
\]

So, the answers are:

- The 43rd term \( a_{43} = 174 \)
- The sum of the first 12 terms \( S_{12} = 336 \)