If `a_(1),a_(2),a_(3),….` are in A.P., then `a_(p),a_(q),a_(r)` are in A.P. if p,q,r are in

Options

A. A.P
B. G.P
C. H.P
D. none of these

Explanation

To solve the problem, we need to show that if \( a_1, a_2, a_3, \ldots \) are in Arithmetic Progression (A.P.), then \( a_p, a_q, a_r \) are also in A.P. if \( p, q, r \) are in A.P.

### Step-by-Step Solution:

1. **Understanding A.P.**:
By definition, a sequence \( a_1, a_2, a_3, \ldots \) is in A.P. if the difference between consecutive terms is constant. This means:
\[ a_2 - a_1 = d \quad \text{and} \quad a_3 - a_2 = d \]
where \( d \) is the common difference.

2. **General Form of A.P.**:
The general term of an A.P. can be expressed as:
\[ a_n = a_1 + (n-1)d \]
Therefore, we can write:
\[ a_p = a_1 + (p-1)d, \quad a_q = a_1 + (q-1)d, \quad a_r = a_1 + (r-1)d \]

3. **Condition for \( p, q, r \) to be in A.P.**:
For \( p, q, r \) to be in A.P., the condition is:
\[ 2q = p + r \]

4. **Expressing \( a_p, a_q, a_r \)**:
Now substituting the expressions for \( a_p, a_q, a_r \):
\[ a_p = a_1 + (p-1)d \]
\[ a_q = a_1 + (q-1)d \]
\[ a_r = a_1 + (r-1)d \]

5. **Finding the common difference**:
To show that \( a_p, a_q, a_r \) are in A.P., we need to show:
\[ a_q - a_p = a_r - a_q \]
Calculating \( a_q - a_p \):
\[ a_q - a_p = \left( a_1 + (q-1)d \right) - \left( a_1 + (p-1)d \right) = (q - p)d \]

Now calculating \( a_r - a_q \):
\[ a_r - a_q = \left( a_1 + (r-1)d \right) - \left( a_1 + (q-1)d \right) = (r - q)d \]

6. **Setting the equations equal**:
We need to show:
\[ (q - p)d = (r - q)d \]
Dividing both sides by \( d \) (assuming \( d \neq 0 \)):
\[ q - p = r - q \]
Rearranging gives:
\[ 2q = p + r \]
This confirms that \( p, q, r \) are in A.P.

### Conclusion:
Thus, we have shown that if \( a_1, a_2, a_3, \ldots \) are in A.P., then \( a_p, a_q, a_r \) are also in A.P. if \( p, q, r \) are in A.P.