Question

The 16th term of an Ap is 93 while the 25th term is 173 find the
19th term
Sum of 13th and 56th term
Product of 6th and 13th term

Answer 1

To solve the problem, we need to remember the formula for the \(n\)th term of an arithmetic progression (AP), which is given by:

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

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

We are given:

- The 16th term \(a_{16} = 93\)
- The 25th term \(a_{25} = 173\)

Using the formula for the 16th term:

\[
a + 15d = 93 \quad \text{(1)}
\]

Using the formula for the 25th term:

\[
a + 24d = 173 \quad \text{(2)}
\]

Next, we will subtract equation (1) from equation (2) to eliminate \(a\):

\[
(a + 24d) - (a + 15d) = 173 - 93
\]
\[
24d - 15d = 80
\]
\[
9d = 80
\]
\[
d = \frac{80}{9}
\]

Now that we have \(d\), we can substitute \(d\) back into either equation to find \(a\). We'll use equation (1):

\[
a + 15\left(\frac{80}{9}\right) = 93
\]
\[
a + \frac{1200}{9} = 93
\]
\[
a = 93 - \frac{1200}{9}
\]
Converting \(93\) to a fraction with a denominator of \(9\):

\[
93 = \frac{837}{9}
\]
Now substituting in:

\[
a = \frac{837}{9} - \frac{1200}{9} = \frac{837 - 1200}{9} = \frac{-363}{9} = -\frac{121}{3}
\]

Now, we have \(a = -\frac{121}{3}\) and \(d = \frac{80}{9}\).

### Finding the 19th Term

Now we can find the 19th term \(a_{19}\):

\[
a_{19} = a + (19-1)d = a + 18d
\]
\[
a_{19} = -\frac{121}{3} + 18\left(\frac{80}{9}\right)
\]
Calculating \(18 \times \frac{80}{9}\):

\[
18 \times \frac{80}{9} = \frac{1440}{9}
\]

Now substituting back:

\[
a_{19} = -\frac{121}{3} + \frac{1440}{9}
\]

Converting \(-\frac{121}{3}\) to have the same denominator as \(\frac{1440}{9}\):

\[
-\frac{121}{3} = -\frac{363}{9}
\]
Thus,

\[
a_{19} = -\frac{363}{9} + \frac{1440}{9} = \frac{1440 - 363}{9} = \frac{1077}{9}
\]

### Sum of the 13th and 56th Terms

Finding the 13th term \(a_{13}\):

\[
a_{13} = a + 12d = -\frac{121}{3} + 12\left(\frac{80}{9}\right) = -\frac{121}{3} + \frac{960}{9}
\]
Converting \(-\frac{121}{3}\) again:

\[
-\frac{121}{3} = -\frac{363}{9}
\]
So,

\[
a_{13} = -\frac{363}{9} + \frac{960}{9} = \frac{960 - 363}{9} = \frac{597}{9}
\]

Now finding the 56th term \(a_{56}\):

\[
a_{56} = a + 55d = -\frac{121}{3} + 55\left(\frac{80}{9}\right) = -\frac{121}{3} + \frac{4400}{9}
\]
Converting \(-\frac{121}{3}\):

\[
a_{56} = -\frac{363}{9} + \frac{4400}{9} = \frac{4400 - 363}{9} = \frac{4037}{9}
\]

Now calculating the sum:

\[
a_{13} + a_{56} = \frac{597}{9} + \frac{4037}{9} = \frac{4634}{9}
\]

### Product of the 6th and 13th Terms

Finding the 6th term \(a_{6}\):

\[
a_{6} = a + 5d = -\frac{121}{3} + 5\left(\frac{80}{9}\right) = -\frac{121}{3} + \frac{400}{9}
\]
Converting \(-\frac{121}{3}\):

\[
-\frac{121}{3} = -\frac{363}{9}
\]

Thus,

\[
a_{6} = -\frac{363}{9} + \frac{400}{9} = \frac{400 - 363}{9} = \frac{37}{9}
\]

Now we already have \(a_{13} = \frac{597}{9}\).

Calculating the product:

\[
a_{6} \times a_{13} = \frac{37}{9} \times \frac{597}{9} = \frac{37 \times 597}{81}
\]

Calculating \(37 \times 597 = 22089\).

Thus,

\[
\text{Product} = \frac{22089}{81}
\]

### Summary of Results

1. The 19th term is \(\frac{1077}{9}\).
2. The sum of the 13th and 56th terms is \(\frac{4634}{9}\).
3. The product of the 6th and 13th terms is \(\frac{22089}{81}\).