Question

The sum of 16th term of an ap is -504,while the sum of the 9th term is -126.find the sum of its 30 term

Answer 1

To solve this problem, we need to use the properties of arithmetic progressions (AP).

Let \( S_n \) be the sum of the first \( n \) terms of an AP, which can be expressed as:

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

where \( a \) is the first term and \( d \) is the common difference.

We are given:

- \( S_{16} = -504 \)
- \( S_{9} = -126 \)

Using the formula for the sum of the first \( n \) terms, we can write:

1. For \( n = 16 \):

\[
S_{16} = \frac{16}{2} \times (2a + 15d) = 8(2a + 15d) = -504
\]

This simplifies to:

\[
2a + 15d = \frac{-504}{8} = -63 \quad \text{(1)}
\]

2. For \( n = 9 \):

\[
S_{9} = \frac{9}{2} \times (2a + 8d) = \frac{9}{2}(2a + 8d) = -126
\]

Thus, we have:

\[
2a + 8d = \frac{-126 \times 2}{9} = -28 \quad \text{(2)}
\]

Now, we need to solve the two equations (1) and (2):

From equation (1):

\[
2a + 15d = -63
\]

From equation (2):

\[
2a + 8d = -28
\]

Subtract equation (2) from equation (1):

\[
(2a + 15d) - (2a + 8d) = -63 - (-28)
\]

This simplifies to:

\[
7d = -35
\]

Solving for \( d \):

\[
d = \frac{-35}{7} = -5
\]

Now substitute \( d = -5 \) back into equation (2) to find \( 2a \):

\[
2a + 8(-5) = -28
\]

Simplifying:

\[
2a - 40 = -28
\]

Adding 40 to both sides:

\[
2a = 12
\]

Thus,

\[
a = 6
\]

Now we have \( a = 6 \) and \( d = -5 \).

We need to find the sum of the first 30 terms, \( S_{30} \):

\[
S_{30} = \frac{30}{2} \times (2a + 29d)
\]

Simplifying \( S_{30} \):

\[
S_{30} = 15 \times (2 \cdot 6 + 29 \cdot -5)
\]

Calculating inside the parentheses:

\[
2 \cdot 6 = 12
\]
\[
29 \cdot -5 = -145
\]

Thus:

\[
S_{30} = 15 \times (12 - 145) = 15 \times (-133) = -1995
\]

So, the sum of the first 30 terms is:

\[
\boxed{-1995}
\]