Question

Given that \( n \) number of arithmetic means are inserted between two pairs of numbers \( a,2b \) and \( 2a,b \); where \( a,b \in R \). If the \( m^{th} \) means in the two cases are the same, then the ratio \( a:b \) is equal to

• A. \( m : (n-m+1) \)
• B. \( n : (n-m+1) \)
• C. \( (n-m+1) : m \)
• D. \( (n-m+1) : n \)

Hint:

In A.P. problems, use \( T_k = a + (k-1)d \) or direct formula of inserted meansCarefully equate corresponding terms.

Answer 1

The Correct Option is A

Step 1: Use formula of A.P.
If \( n \) arithmetic means are inserted between two numbers, total terms become \( n+2 \).

Step 2: First case \( a \) and \( 2b \).
Common difference:
\[ d_1 = \frac{2b-a}{n+1} \]
\( m^{th} \) mean is:
\[ T_m = a + m\cdot \frac{2b-a}{n+1} \]

Step 3: Second case \( 2a \) and \( b \).
Common difference:
\[ d_2 = \frac{b-2a}{n+1} \]
\( m^{th} \) mean is:
\[ T_m = 2a + m\cdot \frac{b-2a}{n+1} \]

Step 4: Equate both means.
\[ a + m\frac{2b-a}{n+1} = 2a + m\frac{b-2a}{n+1} \]

Step 5: Multiply both sides by \( n+1 \).
\[ a(n+1) + m(2b-a) = 2a(n+1) + m(b-2a) \]

Step 6: Simplify equation.
\[ an + a + 2mb - ma = 2an + 2a + mb - 2ma \]
Rearranging:
\[ mb = a(n-m+1) \]

Step 7: Write ratio.
\[ \frac{a}{b} = \frac{m}{n-m+1} \]
\[ \boxed{m : (n-m+1)} \]