Let `a_1,a_2` ,…. , `a_10` be in A.P. and `h_1,h_2` …. `h_10` be in H.P. If `a_1=h_1=2` and `a_10 = h_10 =3` , then `a_4 h_7` is :

Options

A. 2
B. 3
C. 5
D. 6

Answer

To solve the problem step by step, we will first find the terms of the arithmetic progression (A.P.) and then the harmonic progression (H.P.) based on the given conditions.

### Step 1: Identify the terms of the A.P.
We know that:
- \( a_1 = 2 \)
- \( a_{10} = 3 \)

The general formula for the \( n \)-th term of an A.P. is given by:
\[ a_n = a_1 + (n-1)d \]
where \( d \) is the common difference.

For \( n = 10 \):
\[ a_{10} = a_1 + 9d = 3 \]
Substituting \( a_1 = 2 \):
\[ 2 + 9d = 3 \]
Solving for \( d \):
\[ 9d = 3 - 2 \\ 9d = 1 \\ d = \frac{1}{9} \]

### Step 2: Find \( a_4 \)
Now we can find \( a_4 \):
\[ a_4 = a_1 + 3d = 2 + 3 \times \frac{1}{9} = 2 + \frac{3}{9} = 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3} \]

### Step 3: Identify the terms of the H.P.
The first term of the H.P. is:
- \( h_1 = 2 \)
- \( h_{10} = 3 \)

The \( n \)-th term of a H.P. can be expressed in terms of the corresponding A.P. of the reciprocals:
\[ \frac{1}{h_n} = \frac{1}{h_1} + (n-1)d' \]
where \( d' \) is the common difference of the A.P. formed by the reciprocals of the H.P. terms.

For \( n = 10 \):
\[ \frac{1}{h_{10}} = \frac{1}{2} + 9d' \]
Substituting \( h_{10} = 3 \):
\[ \frac{1}{3} = \frac{1}{2} + 9d' \]
Solving for \( d' \):
\[ 9d' = \frac{1}{3} - \frac{1}{2} \]
Finding a common denominator (which is 6):
\[ \frac{1}{3} = \frac{2}{6}, \quad \frac{1}{2} = \frac{3}{6} \\ 9d' = \frac{2}{6} - \frac{3}{6} = -\frac{1}{6} \\ d' = -\frac{1}{54} \]

### Step 4: Find \( h_7 \)
Now we can find \( h_7 \):
\[ \frac{1}{h_7} = \frac{1}{h_1} + 6d' = \frac{1}{2} + 6 \left(-\frac{1}{54}\right) \]
Calculating:
\[ \frac{1}{h_7} = \frac{1}{2} - \frac{6}{54} = \frac{1}{2} - \frac{1}{9} \]
Finding a common denominator (which is 18):
\[ \frac{1}{2} = \frac{9}{18}, \quad \frac{1}{9} = \frac{2}{18} \\ \frac{1}{h_7} = \frac{9}{18} - \frac{2}{18} = \frac{7}{18} \]
Thus:
\[ h_7 = \frac{18}{7} \]

### Step 5: Calculate \( a_4 \times h_7 \)
Now we can find \( a_4 \times h_7 \):
\[ a_4 \times h_7 = \frac{7}{3} \times \frac{18}{7} \]
The \( 7 \) cancels out:
\[ = \frac{18}{3} = 6 \]

### Final Answer
Thus, the value of \( a_4 \times h_7 \) is:
\[ \boxed{6} \]