Question

In an Ap the sum of the first six terms is 21 and the seventh term is three times the sum of third and fourth term. Find the first term and common difference

Answer 1

In an arithmetic progression (AP), the sum of the first \( n \) terms is given by the formula:

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

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

Given that the sum of the first six terms \( S_6 = 21 \), we can express this as:

\[
S_6 = \frac{6}{2} \times (2a + 5d) = 21
\]

Simplifying this gives:

\[
3(2a + 5d) = 21
\]

Dividing both sides by 3:

\[
2a + 5d = 7 \quad \text{(1)}
\]

We also know that the seventh term \( a + 6d \) is three times the sum of the third term and the fourth term. The third term is \( a + 2d \) and the fourth term is \( a + 3d \), so their sum is:

\[
(a + 2d) + (a + 3d) = 2a + 5d
\]

According to the problem statement:

\[
a + 6d = 3(2a + 5d)
\]

Expanding the right side gives:

\[
a + 6d = 6a + 15d
\]

Rearranging this, we can subtract \( a \) and \( 6d \) from both sides:

\[
0 = 5a + 9d
\]

This can be rearranged to:

\[
5a + 9d = 0 \quad \text{(2)}
\]

Now we have the system of equations:

1. \( 2a + 5d = 7 \)
2. \( 5a + 9d = 0 \)

From equation (2), we can express \( a \) in terms of \( d \):

\[
5a = -9d \quad \Rightarrow \quad a = -\frac{9}{5}d \quad \text{(3)}
\]

Substituting equation (3) into equation (1):

\[
2\left(-\frac{9}{5}d\right) + 5d = 7
\]

This simplifies to:

\[
-\frac{18}{5}d + 5d = 7
\]

To combine the terms, convert \( 5d \) into a fraction:

\[
-\frac{18}{5}d + \frac{25}{5}d = 7
\]

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

Now, multiply both sides by 5:

\[
7d = 35 \quad \Rightarrow \quad d = 5
\]

Now, substitute \( d = 5 \) back into equation (3) to find \( a \):

\[
a = -\frac{9}{5} \cdot 5 = -9
\]

Thus, the first term \( a \) and the common difference \( d \) are:

\[
\boxed{-9} \quad \text{and} \quad \boxed{5}
\]