Positive Series
− 2 3 , − 2 0 , − 1 7 , − 1 4 , …
The above are the first few terms of an arithmetic progression .
If the sum of the first n terms of this progression is positive, what is the smallest possible value of n ?
The answer is 17.
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4 solutions
Let the first term of sequence be denoted by a , a = − 2 3 and let the common difference be denoted by d ,
d = T n − T n − 1 = T n − 1 − T n − 2 = ⋯ = T 2 − T 1 = + 3 .
Sum of first n terms of the sequence is given by S n = 2 n ( 2 a + ( n − 1 ) d )
S n = 2 n { 2 a + ( n − 1 ) d } 2 n { − 4 6 + ( n − 1 ) 3 } 3 n − 4 9 n min > 0 > 0 > 0 Since n ∈ N = 1 7
Nice.
Small mistake: It should be n ∈ N , not n ∈ I + because I + is not valid/predefined math notation.
Answer for n is actually 16.33.... We take n = 1 7 but not 1 6 ...Why is that? I couldn't understand.
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Actually, if you check the solution set for the quadratic inequality
S n = 2 n { 2 a + ( n − 1 ) d } > 0 2 n { − 4 6 + ( n − 1 ) 3 } > 0 2 3 n 2 − 4 9 n > 0 n ( 3 n − 4 9 ) > 0
You'll observe that the range of values of n comes out to be ( − ∞ , 0 ) ∪ ( 1 6 . 3 3 , ∞ ) . Since the first range is unacceptable as we can have only positive integral values for n therefore, in the second range, the smallest value of n comes out to be 1 7 .
Thought we would have n( 3n - 49) > 0, so why is 3n - 49 > 0 when we know that n > 0 generally for a sequence? Viewed it from a quadratic inequality approach.
The average of the first n terms must be slightly more than zero, so the n th term will be slightly more than + 2 3 . That means we need more than 3 2 3 − ( − 2 3 ) = 1 5 3 1 terms. The average of the first 16 terms would be − 2 3 + 2 1 5 ⋅ 3 = − 2 1 , which is slightly too small; the average of the first 17 terms would be − 2 3 + 2 1 6 ⋅ 3 = 1 . Thus n = 1 7 is the answer.
Implicitly I used: Lemma: The average of the first n terms of an arithmetic progression is equal to the average of the first and the n th term, ⟨ a i ⟩ 1 … n = 2 a 1 + a n = a 1 + 2 n − 1 ⋅ d , where d is the increment.
Using this equation, we can also derive immediately that n = 1 + d 2 ( ⟨ a i ⟩ − a 1 ) and the condition ⟨ a i ⟩ > 0 gives n > 1 + 3 2 ( 0 − ( − 2 3 ) ) = 1 + 3 4 6 = 1 6 3 1 , from which the answer follows directly.
Very creative!!
U can use formulae but I'm so lazy I've used easy method that count ask -ve terms and double it to cancel it and again add 1 to it to become it +ve from 0 USE SIMPLE METHOD TO SOLVE PROBLEMS
As a quick mental method, it is quick to find that the 8th term is the last negative term and equal to -2. The first positive term follows which is 1.
As the first 8 positive terms (reversing their order) are each one less than the absolutely value of the first 8 terms, we can quickly deduce that the sum of the first 16 terms will be -8 and the 17th term will be big enough to overcome this deficit...
Hence the answer is 17.