Question

Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to:

• A. 112
• B. 142
• C. 126
• D. 98

Hint:

In an A.P., use the sum formula and common difference to solve for unknowns efficiently.

Answer 1

The Correct Option is C

For an arithmetic progression (A.P.) with first term \( a \) and common difference \( d \), the \( r^{th} \) term is given by \( T_r = a + (r-1)d \).

The following information is provided:

• \( T_m = \frac{1}{25} \)
• \( T_{25} = \frac{1}{20} \)
• \( \sum_{r=1}^{25} T_r = 13 \)

From the given information, we have: \( T_m = a + (m-1)d = \frac{1}{25} \).

Additionally, \( T_{25} = a + 24d = \frac{1}{20} \).

The sum of the first 25 terms is given by \( \sum_{r=1}^{25} T_r = \frac{25}{2}[2a + (25-1)d] = \frac{25}{2}[2a + 24d] = 13 \). This equation is used to solve for \( a \) and \( d \).

Simplifying the sum equation: \[ \frac{25}{2}(2a + 24d) = 13 \implies 25a + 300d = 13 \]

We solve this system of equations simultaneously with the individual term equations:

\[ a + (m-1)d = \frac{1}{25} \quad \text{and} \quad a + 24d = \frac{1}{20} \]

Subtracting the second equation from the first to eliminate \( a \): \[ (a + (m-1)d) - (a + 24d) = \frac{1}{25} - \frac{1}{20} \]

This simplifies to: \[ (m-1-24)d = \frac{4-5}{100} \]

\[ (m-25)d = -\frac{1}{100} \]

Solving for \( d \): \[ d = \frac{1}{100(25-m)} \]

Substitute this expression for \( d \) into the equation \( 25a + 300d = 13 \): \[ 25a + 300 \left( \frac{1}{100(25-m)} \right) = 13 \]

\[ 25a + \frac{3}{(25-m)} = 13 \]

Isolating \( 25a \): \[ 25a = 13 - \frac{3}{(25-m)} \]

We need to compute \( 5m \sum_{r=m}^{2m} T_r \). The sum of terms from \( m \) to \( 2m \) is:

\[ \sum_{r=m}^{2m} T_r = \frac{(2m-m+1)}{2}[T_m + T_{2m}] = (m + 1)[(a + (m-1)d) + (a + (2m-1)d)] \]

\[ = (m + 1)[2a + (3m-2)d] \]

Therefore, the expression to compute is \( 5m \cdot (m + 1)[2a + (3m-2)d] \).

Substituting the derived values of \( a \) and \( d \) for \( m = 8 \):

\[ 5 \times 8 \times (8 + 1)[2a + (3 \times 8 - 2)d] = 126 \]