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22 Apr 2020, 02:34
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Average of an evenly spaced sequence a,b,c,d,e (where a-e are evenly spaced terms) is always a+e/2 = b+d/2=c for eg consider a sequence 1,2,3,4,5 AM= 5+1/2=2+4/2=3
ques: What is the average (arithmetic mean) of eleven consecutive integers?
let the sequence be x,x+1,x+2....x+10
st 1: x (first term) + (x+8) (9th term)/2 = 7
therefore x = 3 we can find the AM hence sufficient
st 2: (x+2)(third term) + (x+10) (11th term)/2=9
x=3 we can find the AM hence sufficient
answer D
13 Sep 2021, 21:03
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Simply, the average of 11 consecutive integers will be the middle (median) integer, because the set is evenly spaced and there are an odd number of terms in the set. The first condition gives us the middle (median) of the first nine, so we know the rest of the set, and the second condition gives us the middle integer for the last nine, so we know the remaining numbers in the set.
14 Sep 2021, 08:36
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What is the average (arithmetic mean) of eleven consecutive integers?
Let's consider 5 consecutive integers starting from 3.
3,4,5,6,7
Average = sum/n
Avg = 25/5 = 5 i.e the 3 rd term in the series.
if the no of terms is odd and the series is based on consecutive integers or equally spaced terms, then average = (n+1)/2 th term
Here , the average = (n+1)/2 th term = (5+1)/2 = 3 rd term = 5.
if the no of terms is even and the series is based on consecutive integers or equally spaced terms, then the average will be average of (n/2) th term and (n/2 +1) th term.
In this question, we have 11 consecutive integers i.e no of terms is odd.
So average = (n+1)/2 th term = (11+1)/2 th term = 6 th term in the series. So in order to find the average of these terms, we need to find the 6th term in the series.
(1) The average of the first nine integers is 7
No of terms is 9 i.e odd and they are consecutive integers. The average = (9 +1)/2 the term = 5 th term
Since average is given as 7, we can say that 5 th term in the series is 7.
So the 6 th term will be 8 as they are consecutive integers.
Since we are able to find the 6 th term which is apparently the average of the series, Statement 1 alone is sufficient.
(2) The average of the last nine integers is 9
We can apply the same logic we used in statement 1 . i.e 5 th term = 9
So the 6 th term will be 10.
Statement 2 alone is sufficient to get the average of the numbers.
Option D is the answer.
Thanks,
Clifin J Francis,
GMAT SME
Let's consider 5 consecutive integers starting from 3.
3,4,5,6,7
Average = sum/n
Avg = 25/5 = 5 i.e the 3 rd term in the series.
if the no of terms is odd and the series is based on consecutive integers or equally spaced terms, then average = (n+1)/2 th term
Here , the average = (n+1)/2 th term = (5+1)/2 = 3 rd term = 5.
if the no of terms is even and the series is based on consecutive integers or equally spaced terms, then the average will be average of (n/2) th term and (n/2 +1) th term.
In this question, we have 11 consecutive integers i.e no of terms is odd.
So average = (n+1)/2 th term = (11+1)/2 th term = 6 th term in the series. So in order to find the average of these terms, we need to find the 6th term in the series.
(1) The average of the first nine integers is 7
No of terms is 9 i.e odd and they are consecutive integers. The average = (9 +1)/2 the term = 5 th term
Since average is given as 7, we can say that 5 th term in the series is 7.
So the 6 th term will be 8 as they are consecutive integers.
Since we are able to find the 6 th term which is apparently the average of the series, Statement 1 alone is sufficient.
(2) The average of the last nine integers is 9
We can apply the same logic we used in statement 1 . i.e 5 th term = 9
So the 6 th term will be 10.
Statement 2 alone is sufficient to get the average of the numbers.
Option D is the answer.
Thanks,
Clifin J Francis,
GMAT SME
