If z1, z2, z3, ..., zn is a series of zn is a series of n consecutive

z1, z2, z3, ..., zn is a series of n consecutive even integers.

(1) Because the median of the series (4n) is even, n must be odd (n=3,5,7...).

Sum of all n elements: ... +(4n-4) +(4n-2) +4n +(4n+2) +(4n+4) +... = n*4n = 4n^2
Thus, the sum of the integers in this series is a perfect square (4n^2)

SUFFICIENT

(2) zn = z1 +12 = z1 +2*(n-1) --> n=7
z1=2, then z1+...+z7 = 56 (not a perfect square)
z1=22, then z1+...+z7 = 196 (a perfect square)

We don't know value of z1 or z7. Thus, we cannot deduce whether the sum of the integers in this series is a perfect square.

NOT SUFFICIENT

FINAL ANSWER IS (A)

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sum n/2(2a+2(n-1) = perf square?

(1) sufic
med arith progression is the avg
4n=avg=sum/n, 4n^2=sum, 2^2n^2=perfsqr.

(2) insufic
tn-a=12, a+2(n-1)-a=12, 2n=14, n=7;
no info about 'a'

Ans (A)

If \(z_1, z_2, z_3, ..., z_n\) is a sequence of n consecutive even integers, is the sum of the integers in this series a perfect square?

Since a integers are even the sum would be even. So, we are looking for even-perfect square - 4, 16, 36, 64 ... etc.
Note that if n is even, then mean = \(\frac{sum-of-two-middle-even-integers}{2}\) = some odd integer
Then, Sum of the integers = mean*even number of integers = odd*even
Thus, it would invalidate the question as
Perfect square = even*even

Hence n is always odd.

(1) The median of the series is equal to 4n
Since n is odd median would be mean.
Sum = mean * number of integers = 4n*n = \((2n)^2\)

Hence, the sum would always be a perfect square.
Example:
If n = 3, 4n = 12. Series has 10, 12 and 14 that sum 36. YES
If n = 7, 4n = 28. Series has 22, 24, 26, 28, 30, 32, 34 that sum to 28*7 = 14*14 YES

SUFFICIENT.

(2) \(z_n - z_1 = 12\)
Because \(\frac{12}{2} = 6\)
Thus total integers from \(z_1\) to \(z_n\) including them are 6+1 = 7, n = 7
But it is not defined which 7 even integers.
If the numbers are 2,4, 6, 8, 10, 12, 14 then sum = 8*7 = 56 NO
If the integers are 22, 24, 26, 28, 30, 32, 34 then sum = 28*7 = 14*14 YES

INSUFFICIENT.

Answer A.

Solution

Step 1: Analyse Question Stem

• \(z_1, z_2, z_3, …..,z_n\) is a sequence of n consecutive integers.

o So, the above sequence can be written as \(2k, 2k+2 , 2k+4 ,……..,2k+2*(n-1)\)

 Where k is an integer.
• Let us assume that S is the sum of the integers in the given sequence.

o So, \(S = z_1 + z_2 +z_3 +……+z_n = (2k) + (2k + 2) + (2k + 4) …..+(2k +2*(n-1)) = \frac{(n)}{2}*[2k + 2k+2*(n-1)]\)
o Thus, \(S= n [n + 2k -1] … Eq.(1)\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

Statement 1: The median of the series is equal to 4n

• Median of the series = (first term + last term)/2 \(= 4n \)

o \(⟹ \frac{(2k + 2k +2*(n-1))}{2} = 4n \)
o \(⟹2k + n – 1 = 4n ⟹ n = 4n – 2k + 1\)

 This means n is odd.
• Thus, mean of the series = median of the series \(= 4n ⟹ \frac{S}{ n} = 4n ⟹ S = 2^2*n^2\), which is a perfect square.

Hence, statement 1 is sufficient and we can eliminate answer Options B, C and E.
Statement 2: \(z_n – z_1 = 12\)

• \(z_n = z_1 + 12\)
• \(2k +2*(n-1) = 2k + 12 ⟹ n = 7\)
• Substituting, the value of n in the equation (1), we get

o \(S = 7*(6+2k)\)

However, we don’t know k so we cannot say if S is a perfect square.
Thus, the correct answer is Option A.

This is a “Yes-No” type of DS question. We know that \(z_1\), \(z_2\), \(z_3\), \(z_4\)….. is a sequence of even consecutive integers. We are trying to find out if the sum of all these integers will yield a perfect square.

From statement I alone, the median of the series is 4n.

The median of any set of data values is that value that divides the data set into two equal halves. It depends on the number of values in the ordered data set.
Since the question says z1, z2, z3… zn are consecutive even integers, it’s already in order.

The best way to deal with statement I is to take some values for n and assume the general form of even integers.
Let \(z_1\) = 2k, \(z_2\) = 2k + 2, \(z_3\) = 2k +4 and so on.

If n is even, say n = 4, the terms would be \(z_1\), \(z_2\), \(z_3\), and \(z_4\) and the median would be \(\frac{z_2 + z_3 }{ 2}\) = 4*4 = 16 OR \(z_2\) + \(z_3\) = 32.

\(z_2\) = 2k + 2 and \(z_3\) = 2k + 4. Therefore, 4k + 6 = 32 or 4k = 26 which gives us k = 6.5. Clearly this is impossible since ‘k’ has to be an integer.
This helps us rule out the possibility that the number of terms is even. The number of terms i.e. the value of ‘n’ HAS TO be odd.

Let n = 1, \(z_1\) = Median = 4. The sum of the integers IS a perfect square.

Let n = 3. \(z_1\), \(z_2\) and \(z_3\) are the terms with z2 being the median; Therefore, 2k + 2 = 12 or 2k = 10 or k = 5. Also, since the terms are consecutive, mean = median = 12.
Sum of terms = Mean * number of terms = 12 * 3 = 36. The sum is a perfect square.

Let n = 5. \(z_3\) = 2k + 4 = Median = 20. Therefore, k = 8.
Sum of terms = 20 * 5 = 100. The sum is a perfect square.

From statement I alone, we can conclude that the sum of the series is a perfect square. Statement I alone is sufficient. Answer options B, C and E can be eliminated. Possible answers are A or D.

From statement II alone, \(z_n\) – \(z_1\) = 12. This means \(z_n\) = \(z_1\) + 12. This helps us understand that the sequence has 7 consecutive even integers.

However, knowing the number of integers is not sufficient to say if the sum of these integers will be a perfect square.

For example, if \(z_1\) = 2 and \(z_7\) = 14, mean = 8 and sum of terms = 8 * 7 = 56, not a perfect square.
On the other hand, if \(z_1\) = 22 and \(z_7\) = 32, mean = 28 and sum of terms = 28 * 7 = 196 which IS a perfect square.

Statement II alone is insufficient to answer the question. Answer option D can be eliminated.
The correct answer option is A.

Hope that helps!

Given,
n consecutive even integers. Sum = 2 x n/2 (N+1).

St 1. Speaks of median.
Insufficient
St 2, Zn-Z1= 12, i.e, nos can be 2,4,8,10,12,14, Sum =50, or 4,8,10,12,14,16 , Sum =70,
Sum is not perfect square,

If z1,z2,z3, ..., zn is a sequence of n consecutive even integers, is the sum of the integers in this series a perfect square?

(1) The median of the series is equal to 4n

(2) zn - z1 = 12
stem: 2k,2k+2,2k+4,....2k+2(n-1)
1) 2k+2k+2(n-1)/2=4n
2(n-1)=4k
insufficient
2) 2k+2(n-1)-2k=12
n=7
still we dont know the value of k
insufficient
1+2), sufficient
Ans C

z1,z2,z3,...zn is a sequence of n consecutive even integers.
1) Median = 4n
If n=1, series is 4, Sum = 4
If n=3, series is 10,12,14, Sum = 36
If n=5, series is 16,18,20,22,24, Sum = 100

Hence sum is always a perfect square.

(Note: for n as even, the statement (1) is not applicable as for even number of consecutive even numbers, the median is odd.)

SUFFICIENT.

(2) zn - z1 = 12
This means there are 7 consecutive even terms in the series.

If terms are -6,-4,-2,0,2,4,6
Sum = 0, a perfect square.

If terms are 2,4,6,8,10,12,14
Sum = 56, not a perfect square.

NOT SUFFICIENT.

Hence, only (1) statement alone is sufficient.
Answer is A.

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If \(z1_, z_2, z_3, ..., z_n\) is a sequence of n consecutive even integers, is the sum of the integers in this series a perfect square?


(Statement1): The median of the series is equal to 4n
Since the sequence has consecutive numbers, mean is equal to median:

Sum = Mean * Number of terms = \(4n*n = (2n)^{2}\)
Always YES.
Sufficient

(Statement2): \(z_n - z_1 = 12\)
\(z_1 + 2(n-1) -z_1= 12\)
\(n=7 \)
---> \(a +(a+2)+ (a+4)+ (a+6)+ (a+8)+ (a+10)+ (a+12)= 7a+ 42\)
Nothing tells us about what \(a\) is.
Insufficient

Answer (A).

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