If the sum of the first 31 terms of an arithmetic progression consisti

Question

If the sum of the first 31 terms of an arithmetic progression consisting of 46 terms is zero, then which of the following MUST be true? (An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant.)

I. The 31st smallest term is zero.

II. The 16th largest term is zero.

III. The sum of the largest and smallest terms of the sequence is positive.

I. 31st smallest term is zero 
II. 16th largest term is zero
III. The sum of the largest and smallest terms of the sequence is positive

A. I only
B. II only
C. III only
D. I and II only
E. None of the above

Bunuel · Accepted Answer

Official Solution:

If the sum of the first 31 terms of an arithmetic progression consisting of 46 terms is zero, then which of the following MUST be true? (An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant.)

I. The 31st smallest term is zero.

II. The 16th largest term is zero.

III. The sum of the largest and smallest terms of the sequence is positive.

I. The 31st smallest term is zero.

II. The 16th largest term is zero.

III. The sum of the largest and smallest terms of the sequence is positive.

A. I only
B. II only
C. III only
D. I and II only
E. None of the above

Firstly, note that we don't know whether the sequence is increasing or decreasing.

If the sequence is increasing, then the fact that the sum of the first 31 terms of the arithmetic progression is zero would imply that:
 
 a_1 \leq ... \leq (a_{16} = 0) \leq ... \leq a_{31} \leq ... \leq a_{46} 
 
If the sequence is decreasing, then the fact that the sum of the first 31 terms of the arithmetic progression is zero would imply that:
 
 a_1 \geq ... \geq (a_{16} = 0) \geq ... \geq a_{31} \geq ... \geq a_{46} 
 
As we can see, none of the options MUST be true.

Answer: E

wadhwakaran · Answer

T1, T2, T3, ------, T31, T32, T33, ------, T46

S31=0
31/2(a+a+30d)=0
2a+30d=0
a+15d=0
meaning thereby that the T16 is zero

I. 31st smallest term is zero  In case of the increasing AP, this will be a possibility that 31st smallest term is zero.  
II. 16th largest term is zero  In case of the decreasing AP, this will be a possibility that 16th largest term is zero. 
III. The sum of the largest and smallest terms of the sequence is positive  In case of increasing AP, this will be positive and in case of decreasing AP, this will be negative

All the above 3 are possibilities but none of them are sure result. 
Therefore option E

sanjitscorps18 · Answer

Consider two arithmetic progressions
P1 is -15, -14,..........0..........14, 15..............30
and
P2 is 15, 14,.........0..........-14, -15,.............-30

In P2 we have the 16th largest term as 0 but that is not true in P1

In P2 we have the 31st smallest term as 0 but that is not true in P1

In P1 the sum of the largest and smallest terms of the sequence is positive
In P2 the sum of the largest and smallest terms of the sequence is negative

None of the options seem to fulfill the MUST true condition hence Option E

kkannan2 · Answer

Hi  Bunuel  could you post the official solution for this question? I am getting the answer to be B since in an AP, mean=median, so the (31+1)/2=16th term should be 0. I'm curious why the answer is E.

mandipsah · Answer

Since there are 46 terms and the sum of first 31 terms is zero, so the two order possible are:

order 1: -15, -14, -13, .............0...............13,14,15,16,..............,29,30
order 2: 15, 14, 13,..................0,..............-13,-14,-15,-16,..................,-29,-30

As we can see, the 16th largest term is 0 in order 2 but not in order 1.
So, B cannot be the answer, so E is the answer. I hope this clarifies your doubts.

siddharth_ · Answer

Can we also consider 0,0,0,0.... (all 46 terms 0 -> An AP with 1st term and common diff = 0) as one of the cases for such questions?

Bunuel · Answer

_____________________
Yes.

shriirah · Answer

How is this question any different from this linked question? https://gmatclub.com/forum/if-the-sum-o ... 98379.html

Kinshook · Answer

If the sum of the first 31 terms of an arithmetic progression consisting 46 terms is zero, then which of the following MUST be true ? (An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant)

Let the arithmetic progression have: -
a - > first term
d - > common difference
Number of terms = 46

Sum of first 31 terms = 31/2 {2a + 30d} = 0
16th term = a + 15d = 0
Sum of largest and smallest terms = a + 45d + a = 2a + 45d = 2(a+15d) + 15d = 15d

Case 1: a>0; d<0
Sum of largest and smallest terms = 15d < 0
16th largest term = a + 15d = 0
31st smallest term = a + 45d - 30d = a + 15d = 0

Case 2: a<0; d>0
Sum of largest and smallest terms = 15d > 0
16th largest term = a + 45d - 15d = a + 30d = 15d > 0
31st smallest term = a + 29d = a + 15d + 14d = 14d > 0

I. 31st smallest term is zero : May or may not be true
II. 16th largest term is zero : May or may not be true
III. The sum of the largest and smallest terms of the sequence is positive : May or may not be true

A. I only
B. II only
C. III only
D. I and II only
E. None of the above

IMO E

Krunaal · Answer

The linked question asks for which statement  could  be true, whereas this question asks for which statement  must  be true.

If the sum of the first 31 terms of an arithmetic progression consisti

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