Q.A man accepts a position with an initial salary of Rs $5200$ per month. It is understood that he will receive an automatic increase of Rs $320$ in the very next month and each month thereafter.

The problem describes a situation where a fixed amount is added to a salary each month, which is the definition of an arithmetic progression (AP).

Concept: Arithmetic Progression (AP)

1. To find the salary for the tenth month: The initial salary is $a_1 = 5200$. The monthly increase is the common difference $d = 320$. The $n$-th term of an AP is given by $a_n = a_1 + (n-1)d$. For the tenth month ($n=10$), $a_{10} = 5200 + (10-1) \times 320 = 5200 + 9 \times 320 = 5200 + 2880 = 8080$.
   ✓Final answer

   His salary for the tenth month is $\boxed{\text{Rs } 8080}$.

2. To find his total earnings during the first year (12 months): The sum of the first $n$ terms of an AP is given by $S_n = \frac{n}{2}[2a_1 + (n-1)d]$. For the first 12 months ($n=12$), $S_{12} = \frac{12}{2}[2 \times 5200 + (12-1) \times 320]$. $S_{12} = 6[10400 + 11 \times 320] = 6[10400 + 3520] = 6[13920] = 83520$.
   ✓Final answer

   His total earnings during the first year are $\boxed{\text{Rs } 83520}$.

The situation described here is a classic example of an Arithmetic Progression (AP). An AP is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference.

In this problem:

• The initial salary is the first term of our sequence.
• The automatic increase each month is the common difference, as the salary increases by the same fixed amount every month.

Understanding this allows us to use the standard formulas for arithmetic progressions to find specific terms (like the salary in a particular month) and the sum of terms (like total earnings over a period).

Let's identify the given values:

• Initial salary ($a$) = Rs $5200$
• Monthly increase ($d$) = Rs $320$

Part (a): Find his salary for the tenth month.

1. Identify the parameters:

   We need to find the salary for the tenth month, which means we are looking for the $10$-th term of the arithmetic progression.

   So, $n = 10$.

   The first term $a = 5200$.

   The common difference $d = 320$.

2. Apply the $n$-th term formula:

   The formula for the $n$-th term is $a_n = a + (n-1)d$.

   Substitute the values for $a$, $d$, and $n$:

   $a_{10} = 5200 + (10-1) \times 320$

   $a_{10} = 5200 + 9 \times 320$

3. Calculate the salary:

   First, calculate the product: $9 \times 320 = 2880$.

   Then, add this to the initial salary:

   $a_{10} = 5200 + 2880$

   $a_{10} = 8080$

   So, his salary for the tenth month will be Rs $8080$.

Part (b): What is his total earnings during the first year?

1. Identify the parameters:

   "Total earnings during the first year" means the sum of his salaries for the first $12$ months.

   So, we need to find the sum of the first $12$ terms, which means $n = 12$.

   The first term $a = 5200$.

   The common difference $d = 320$.

2. Apply the sum of $n$ terms formula:

   The formula for the sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.

   Substitute the values for $a$, $d$, and $n$:

   $S_{12} = \frac{12}{2}[2(5200) + (12-1) \times 320]$

   $S_{12} = 6[10400 + 11 \times 320]$

3. Calculate the total earnings:

   First, calculate the product: $11 \times 320 = 3520$.

   Then, add this to $2a$:

   $S_{12} = 6[10400 + 3520]$

   $S_{12} = 6[13920]$

   Finally, multiply by $6$:

   $S_{12} = 83520$

   So, his total earnings during the first year will be Rs $83520$.