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