Sum of the first hundred numbers common to the arithmetic progression 12, 15, 18, …………… and 17, 21, 25,………. Is

Options

A. 56100
B. 65100
C. 61500
D. 51600

Answer

To find the sum of the first 100 numbers common to the arithmetic progressions (AP) given by 12, 15, 18, ... and 17, 21, 25, ..., we will follow these steps:

### Step 1: Identify the first arithmetic progression (AP)
The first AP is:
- First term (a1) = 12
- Common difference (d1) = 3

The general term of this AP can be expressed as:
\[ a_n = 12 + (n-1) \cdot 3 = 3n + 9 \]

### Step 2: Identify the second arithmetic progression (AP)
The second AP is:
- First term (a2) = 17
- Common difference (d2) = 4

The general term of this AP can be expressed as:
\[ b_m = 17 + (m-1) \cdot 4 = 4m + 13 \]

### Step 3: Set the general terms equal to find common terms
To find the common terms between the two APs, we set the general terms equal:
\[ 3n + 9 = 4m + 13 \]

Rearranging gives:
\[ 3n - 4m = 4 \]

### Step 4: Solve for integers n and m
We can express m in terms of n:
\[ 4m = 3n - 4 \]
\[ m = \frac{3n - 4}{4} \]

For m to be an integer, \( 3n - 4 \) must be divisible by 4.

### Step 5: Find valid values of n
To satisfy \( 3n - 4 \equiv 0 \mod 4 \):
\[ 3n \equiv 4 \mod 4 \]
Since \( 4 \mod 4 = 0 \), we have:
\[ 3n \equiv 0 \mod 4 \]

The multiplicative inverse of 3 modulo 4 is 3, so multiplying both sides by 3 gives:
\[ n \equiv 0 \mod 4 \]

Thus, n can take values: 4, 8, 12, ..., which can be expressed as:
\[ n = 4k \text{ for } k = 1, 2, 3, ... \]

### Step 6: Find the common terms
Substituting \( n = 4k \) into the expression for the common terms:
\[ a_{4k} = 3(4k) + 9 = 12k + 9 \]

### Step 7: Identify the first 100 common terms
The first 100 common terms are:
- 21 (when k=1)
- 33 (when k=2)
- 45 (when k=3)
- ...
- 12k + 9 for k = 1 to 100

### Step 8: Find the common difference
The common difference (d) between the common terms is:
\[ d = (12(k+1) + 9) - (12k + 9) = 12 \]

### Step 9: Calculate the sum of the first 100 common terms
Using the formula for the sum of the first n terms of an AP:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
Where:
- n = 100
- a = 21 (the first term)
- d = 12 (the common difference)

Substituting the values:
\[ S_{100} = \frac{100}{2} \cdot (2 \cdot 21 + (100 - 1) \cdot 12) \]
\[ = 50 \cdot (42 + 99 \cdot 12) \]
\[ = 50 \cdot (42 + 1188) \]
\[ = 50 \cdot 1230 \]
\[ = 61500 \]

### Final Answer:
The sum of the first 100 numbers common to the two arithmetic progressions is **61500**.

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