Let's analyze each reaction:
A. Iodoform Reaction:
Acetaldehyde ($ CH_3CHO $) contains a $ CH_3CO- $ group and thus gives a positive iodoform test.
Acetone ($ CH_3COCH_3 $) also contains a $ CH_3CO- $ group and thus gives a positive iodoform test.
Conclusion: Both acetaldehyde and acetone undergo the iodoform reaction.
B. Cannizzaro Reaction:
The Cannizzaro reaction is given by aldehydes that do not have an α-hydrogen. Acetaldehyde has α-hydrogens and thus does not undergo the Cannizzaro reaction.
Acetone is a ketone and thus cannot undergo the Cannizzaro reaction.
Conclusion: Neither acetaldehyde nor acetone undergoes the Cannizzaro reaction.
C. Aldol Condensation:
The aldol condensation is given by aldehydes and ketones having at least one α-hydrogen. Acetaldehyde has α-hydrogens and thus undergoes aldol condensation.
Acetone also has α-hydrogens and thus undergoes aldol condensation.
Conclusion: Both acetaldehyde and acetone undergo aldol condensation.
D. Tollen's Test:
Tollen's test is given by aldehydes. Acetaldehyde is an aldehyde and thus gives a positive Tollen's test.
Acetone is a ketone and does not give the Tollen's test.
Conclusion: Only acetaldehyde undergoes the Tollen's test.
E. Clemmensen Reduction:
The Clemmensen reduction is used to reduce aldehydes and ketones to alkanes. Both acetaldehyde and acetone undergo Clemmensen reduction.
Conclusion: Both acetaldehyde and acetone undergo Clemmensen reduction.
Final Conclusion:
The reactions that both acetaldehyde and acetone undergo are: Iodoform reaction (A), Aldol condensation (C), and Clemmensen reduction (E).
Final Answer:
The final answer is $ A,\ C\ \text{and}\ E\ \text{only} $.
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: