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

Answer 1

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

In an AP, the \( n \)-th term can be expressed as:

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

Given that we have six terms, we can write the sum of the first six terms \( S_6 \):

\[ S_6 = T_1 + T_2 + T_3 + T_4 + T_5 + T_6 \]

Using the formula for the sum of the first \( n \) terms of an AP:

\[ S_n = \frac{n}{2} \times (T_1 + T_n) \]

For six terms, the formula becomes:

\[ S_6 = \frac{6}{2} \times \left(T_1 + T_6\right) = 3 \times (a + (a + 5d)) = 3 \times (2a + 5d) = 6a + 15d \]

According to the problem, this sum is given as 21:

\[ 6a + 15d = 21 \quad (1) \]

Next, consider the seventh term, which is given as:

\[ T_7 = a + 6d \]

We are told that this seventh term is three times the sum of the third and fourth terms. The third term \( T_3 \) and fourth term \( T_4 \) are given by:

\[ T_3 = a + 2d \] \[ T_4 = a + 3d \]

Calculating the sum of the third and fourth terms:

\[ T_3 + T_4 = (a + 2d) + (a + 3d) = 2a + 5d \]

The condition given in the problem states that:

\[ T_7 = 3 \times (T_3 + T_4) \implies a + 6d = 3(2a + 5d) \]

Expanding the right side gives us:

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

This can be rearranged to:

\[ a + 6d - 6a - 15d = 0 \]

Combining like terms results in:

\[ -5a - 9d = 0 \quad (2) \]

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

\[ 5a = -9d \implies a = -\frac{9}{5}d \]

Now we will substitute \( a \) in terms of \( d \) back into equation (1):

\[ 6\left(-\frac{9}{5}d\right) + 15d = 21 \]

This simplifies to:

\[ -\frac{54}{5}d + 15d = 21 \]

Converting \( 15d \) to have a common denominator with \( -\frac{54}{5}d \):

\[ -\frac{54}{5}d + \frac{75}{5}d = 21 \]

Combining the terms gives:

\[ \frac{21}{5}d = 21 \]

To solve for \( d \), multiply both sides by 5:

\[ 21d = 105 \implies d = 5 \]

Now substitute \( d = 5 \) back into the equation for \( a \):

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

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

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