Question

The 14th term of an AP is 96 while the 25th term 173 find The Product of 6th and 13th term

Answer 1

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

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

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

Given:

- The 14th term \( a_{14} = 96 \)
- The 25th term \( a_{25} = 173 \)

We can set up the equations based on the \( n \)-th term formula:

1. For the 14th term:
\[
a + 13d = 96 \quad \text{(1)}
\]

2. For the 25th term:
\[
a + 24d = 173 \quad \text{(2)}
\]

Now we can subtract equation (1) from equation (2) to find \( d \):

\[
(a + 24d) - (a + 13d) = 173 - 96
\]
\[
11d = 77
\]
\[
d = \frac{77}{11} = 7
\]

Now that we have \( d \), we can substitute it back into either equation (1) or (2) to find \( a \). Let's use equation (1):

\[
a + 13(7) = 96
\]
\[
a + 91 = 96
\]
\[
a = 96 - 91 = 5
\]

Now we have \( a = 5 \) and \( d = 7 \).

Next, we need to find the 6th term and the 13th term:

1. The 6th term \( a_6 \):
\[
a_6 = a + (6-1)d = 5 + 5 \cdot 7 = 5 + 35 = 40
\]

2. The 13th term \( a_{13} \):
\[
a_{13} = a + (13-1)d = 5 + 12 \cdot 7 = 5 + 84 = 89
\]

Now, to find the product of the 6th and 13th terms:

\[
\text{Product} = a_6 \times a_{13} = 40 \times 89
\]

Calculating the product:

\[
40 \times 89 = 3560
\]

Thus, the product of the 6th and 13th terms is \( \boxed{3560} \).