Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant.
  • General term: $a_k = a_1 + (k-1)d$, where $a_1$ is the first term and $d$ is the common difference.
  • Sum of first $k$ terms: $S_k = \frac{k}{2}(2a_1 + (k-1)d)$.
• Middle Terms: For an A.P. with $n$ terms:
  • If $n$ is odd, the middle term is $a_{(n+1)/2}$.
  • If $n$ is even, the middle terms are $a_{n/2}$ and $a_{n/2+1}$.

Step-by-Step Solution

Step 1: Determine the Common Difference ($d$)

We are given the first term $a_1 = 8$. The sum of the first four terms is 50. The first four terms are $a_1, a_2, a_3, a_4$. Using the formula $a_k = a_1 + (k-1)d$, these terms are $a_1$, $a_1+d$, $a_1+2d$, and $a_1+3d$. The sum of these terms is: $a_1 + (a_1+d) + (a_1+2d) + (a_1+3d) = 50$ Combining like terms, we get: $4a_1 + 6d = 50$ Substitute the given value $a_1 = 8$: $4(8) + 6d = 50$ $32 + 6d = 50$ Subtract 32 from both sides: $6d = 18$ Divide by 6 to find $d$: $d = \frac{18}{6} = 3$ The common difference is $d=3$.

Step 2: Determine the Number of Terms ($n$)

We are given that the sum of the last four terms is 170. The last four terms of an A.P. with $n$ terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n$. Using the formula $a_k = a_1 + (k-1)d$, these terms are: $a_{n-3} = a_1 + (n-4)d$ $a_{n-2} = a_1 + (n-3)d$ $a_{n-1} = a_1 + (n-2)d$ $a_n = a_1 + (n-1)d$ The sum of these terms is: $(a_1 + (n-4)d) + (a_1 + (n-3)d) + (a_1 + (n-2)d) + (a_1 + (n-1)d) = 170$ Combining like terms: $4a_1 + ((n-4) + (n-3) + (n-2) + (n-1))d = 170$ $4a_1 + (4n - 10)d = 170$ Substitute the known values $a_1 = 8$ and $d = 3$: $4(8) + (4n - 10)(3) = 170$ $32 + 12n - 30 = 170$ $2 + 12n = 170$ Subtract 2 from both sides: $12n = 168$ Divide by 12 to find $n$: $n = \frac{168}{12} = 14$ The A.P. has 14 terms.

Step 3: Identify the Middle Two Terms

Since $n=14$ is an even number, the A.P. has two middle terms. These are the $(n/2)$-th term and the $(n/2 + 1)$-th term. For $n=14$, these are the $14/2 = 7$-th term ($a_7$) and the $(14/2 + 1) = 8$-th term ($a_8$). We calculate these terms using $a_1=8$ and $d=3$: $a_7 = a_1 + (7-1)d = 8 + 6(3) = 8 + 18 = 26$ $a_8 = a_1 + (8-1)d = 8 + 7(3) = 8 + 21 = 29$ The middle two terms are 26 and 29.

Step 4: Calculate the Product of the Middle Two Terms

We need to find the product of $a_7$ and $a_8$. Product = $a_7 \times a_8 = 26 \times 29$ $26 \times 29 = 26 \times (30 - 1) = 26 \times 30 - 26 \times 1 = 780 - 26 = 754$

Common Mistakes & Tips

• Index Errors: Carefully identify the indices of the last four terms ($a_{n-3}, a_{n-2}, a_{n-1}, a_n$) and the middle terms ($a_{n/2}, a_{n/2+1}$).
• Algebraic Simplification: Ensure accuracy when combining like terms and solving for unknowns, especially with the presence of $n$ in the equations for the last four terms.
• Alternative for Last Four Terms: The sum of the last four terms can also be expressed as $4a_n - 6d = 170$. Substituting $a_n = a_1 + (n-1)d$ and solving for $n$ with the known $a_1$ and $d$ can be an alternative approach.

Summary

The problem involves an arithmetic progression where we are given the first term and sums of the first four and last four terms. We first used the sum of the first four terms to find the common difference $d$. Then, using the sum of the last four terms along with the first term and the common difference, we determined the total number of terms $n$. Since $n$ was even, we identified the two middle terms and calculated their values. Finally, we multiplied these two middle terms to obtain the required product.

The final answer is \boxed{754}.