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

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

We are given:

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

We know: \( T_m = a + (m-1)d = \frac{1}{25} \).

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

From \( \sum_{r=1}^{25} T_r = \frac{25}{2}[2a + 24d] = 13 \), we solve for \( a \) and \( d \).

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

Solving simultaneously with previous equations:

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

Eliminate \( a \) by subtracting the two: \[ (m-1)d - 24d = \frac{1}{25} - \frac{1}{20} \]

Calculate: \[ (m-25)d = \frac{4-5}{100} = -\frac{1}{100}\]

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

Substitute \( d \) in \( 25a + 300d = 13 \): \[ 25a + 300 \left( \frac{1}{100(25-m)} \right) = 13 \]

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

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

To compute \( 5m \sum_{r=m}^{2m} T_r \):

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

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

Thus, \( 5m \cdot (m + 1)[2a + (3m-2)d] \):

Upon solving with given \( a \) and \( d \) values:\( m = 8 \) yields:

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