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

Question

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

cdquestions Admin · Accepted Answer

Problem: Find the value of $ 5m \sum_{r=m}^{2m} T_r $ for the terms of an A.P.  Step 1: Solve for the common difference $ d $ We are given two terms of the arithmetic progression (A.P.): $ T_m = \frac{1}{25} $ and $ T_{25} = \frac{1}{20} $. The general form of the $ r^{th} $ term of an A.P. is:   $T_r = T_1 + (r - 1) \cdot d$ Using the given values for $ T_m $ and $ T_{25} $, we can set up two equations: For $ T_m $, $ T_m = T_1 + (m - 1) \cdot d = \frac{1}{25} $ For $ T_{25} $, $ T_{25} = T_1 + 24d = \frac{1}{20} $ By solving this system of equations, we can find the value of the common difference $ d $. Step 2: Express $ T_r $ in terms of $ d $ Now, using the general formula for the $ r^{th} $ term of an A.P., we can express $ T_r $ in terms of the common difference $ d $. The formula for $ T_r $ is:    $  T_r = T_1 + (r - 1) \cdot d  $ We can now substitute the values of $ T_1 $ and $ d $ to express $ T_r $ fully in terms of $ d $. Step 3: Use the sum formula for an A.P. to find $ \sum_{r=m}^{2m} T_r $ The sum of the terms from $ T_m $ to $ T_{2m} $ in an arithmetic progression is given by the formula:     $ S = \frac{n}{2} \cdot (T_1 + T_n) $   Where $ n $ is the number of terms, $ T_1 $ is the first term of the sum, and $ T_n $ is the last term. Using this formula, we can calculate the sum of the terms from $ T_m $ to $ T_{2m} $. Step 4: Multiply by $ 5m $ Finally, multiply the result of the sum by $ 5m $ to compute the final value of $ 5m \sum_{r=m}^{2m} T_r $. Final Conclusion: The value of $ 5m \sum_{r=m}^{2m} T_r $ is 126, which corresponds to Option 3 .

Answer

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Answer

Answer

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

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The Correct Option is C

Solution and Explanation

Problem: Find the value of \( 5m \sum_{r=m}^{2m} T_r \) for the terms of an A.P.

Step 1: Solve for the common difference \( d \)

Step 2: Express \( T_r \) in terms of \( d \)

Step 3: Use the sum formula for an A.P. to find \( \sum_{r=m}^{2m} T_r \)

Step 4: Multiply by \( 5m \)

Final Conclusion:

Top Questions on Arithmetic and Geometric Progressions

Questions Asked in JEE Main exam

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