An arithmetic progression has the property that the 3rd, 6th and 9th terms add up to 84, while the sum of its first 12 terms is 402. Find the sum of its first 4 terms.
Answer and solution
Correct answer: A
Solution
Answer: A — -42
1. Write tn=a+(n−1)d. The three given terms sum to 3a+15d=84, so a+5d=28.
2. S₁₂=6(2a+11d)=402, hence 2a+11d=67.
3. Double the first equation to get₂a+10d=56. Subtract to obtain d=11.
4. Then a=28−55=−27. S₄=2(2a+3d)=2(−54+33)=−42.
5. Check the first four terms: −27,−16,−5,6 sum to −42, option A.
Why this method works:
Terms 3,6,9 are equally spaced in an AP, so their sum is three times the middle termt₆. The mean of 12 terms lies halfway between t₆ andt₇; these two observations determine d without listing the sequence.
Independent check and common trap:
The first four terms−27,−16,−5,6 sum to−42. Also t₃=−5, t₆=28, t₉=61 sum to 84, while t₁₂=94 gives S₁₂=6(−27+94)=402. Negative early terms are allowed because the AP was not restricted to positive numbers.
Apply: t₃+t₆+t₉=3t₆, so t₆=28. S₁₂/12=33.5=(t₆+t₇)/2, giving d=11. Then a=−27 and S₄=−42.
Check: The first four terms−27,−16,−5,6 sum to−42. Also t₃=−5, t₆=28, t₉=61 sum to 84, while t₁₂=94 gives S₁₂=6(−27+94)=402. Negative early terms are allowed because the AP was not restricted to positive numbers.
1. Write tn=a+(n−1)d. The three given terms sum to 3a+15d=84, so a+5d=28.
2. S₁₂=6(2a+11d)=402, hence 2a+11d=67.
3. Double the first equation to get₂a+10d=56. Subtract to obtain d=11.
4. Then a=28−55=−27. S₄=2(2a+3d)=2(−54+33)=−42.
5. Check the first four terms: −27,−16,−5,6 sum to −42, option A.
Why this method works:
Terms 3,6,9 are equally spaced in an AP, so their sum is three times the middle termt₆. The mean of 12 terms lies halfway between t₆ andt₇; these two observations determine d without listing the sequence.
Independent check and common trap:
The first four terms−27,−16,−5,6 sum to−42. Also t₃=−5, t₆=28, t₉=61 sum to 84, while t₁₂=94 gives S₁₂=6(−27+94)=402. Negative early terms are allowed because the AP was not restricted to positive numbers.
G Strategy shortcut
Spot: Arithmetic Progression.Apply: t₃+t₆+t₉=3t₆, so t₆=28. S₁₂/12=33.5=(t₆+t₇)/2, giving d=11. Then a=−27 and S₄=−42.
Check: The first four terms−27,−16,−5,6 sum to−42. Also t₃=−5, t₆=28, t₉=61 sum to 84, while t₁₂=94 gives S₁₂=6(−27+94)=402. Negative early terms are allowed because the AP was not restricted to positive numbers.
G Strategy: Spot: Arithmetic Progression. Apply: t₃+t₆+t₉=3t₆, so t₆=28. S₁₂/12=33.5=(t₆+t₇)/2, giving d=11. Then a=−27 and S₄=−42. Check: The first four terms−27,−16,−5,6 sum to−42. Also t₃=−5, t₆=28, t₉=61 sum to 84, while t₁₂=94 gives S₁₂=6(−27+94)=402. Negative early terms are allowed because the AP was not restricted to positive numbers.