A man serves Rs. 320 in the month of January Rs. 360 in the month of February, Rs. 400 in the month of March. If he continues his saving in the same way:
a. Find his saving in the month of November in the same year.
b. Find his saving in the month of July in the same year.

Answer

To solve the problem step by step, we will analyze the savings of the man for the months given and find the required savings for November and July.

### Step 1: Identify the pattern in the savings
The savings for the first three months are:
- January: Rs. 320
- February: Rs. 360
- March: Rs. 400

We can see that the savings are increasing each month.

### Step 2: Calculate the common difference
To find the common difference (d), we subtract the savings of one month from the next:
- February - January: 360 - 320 = Rs. 40
- March - February: 400 - 360 = Rs. 40

Thus, the common difference \( d = 40 \).

### Step 3: Identify the first term (a)
The first term of the series (savings in January) is:
- \( a = 320 \)

### Step 4: Find the savings for November (11th month)
To find the savings in November, we need to calculate the 11th term of the arithmetic progression (AP). The formula for the nth term of an AP is:
\[ a_n = a + (n - 1) \cdot d \]

Substituting the values:
- \( n = 11 \)
- \( a = 320 \)
- \( d = 40 \)

\[ a_{11} = 320 + (11 - 1) \cdot 40 \]
\[ a_{11} = 320 + 10 \cdot 40 \]
\[ a_{11} = 320 + 400 \]
\[ a_{11} = 720 \]

So, the savings in November will be Rs. 720.

### Step 5: Find the savings for July (7th month)
Next, we calculate the savings in July, which is the 7th term of the AP. Using the same formula:
\[ a_n = a + (n - 1) \cdot d \]

Substituting the values:
- \( n = 7 \)
- \( a = 320 \)
- \( d = 40 \)

\[ a_{7} = 320 + (7 - 1) \cdot 40 \]
\[ a_{7} = 320 + 6 \cdot 40 \]
\[ a_{7} = 320 + 240 \]
\[ a_{7} = 560 \]

So, the savings in July will be Rs. 560.

### Final Answers
a. The savings in the month of November is Rs. 720.
b. The savings in the month of July is Rs. 560.

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