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BuggerinOn
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BuggerinOn
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525

Odd numbers can be represented in the form : n , ( n + 2 )............

So , n + ( n + 2 ) + ( n + 4 ) + ( n + 6 )............+ ( n + 34 ) = 18*534

Or, 18n + 306 = 18*534

Or, 18 ( n + 17 ) = 18*534

Or, n + 17 = 534

So, n = 517

Hence, answer will be (A) 517...
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Here is my solution to this one ->
Let n be the first term
Last term => \(n+17*2=> n+34\)

Now as the series will be in AP.
Mean = Average of the first and the last term
Hence 534 =\(\frac{n+n+34}{2}=> n+17\)
Hence n+17=534=> n=517

Hence A
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Bunuel
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

(A) 517
(B) 518
(C) 519
(D) 521
(E) 525
Sum = n/2[2a+(n-1)d] , a = is the first term , d = is the common difference n=18 as mentioned in the question
Over here
Sum = 18*534 = 9612 , N=18 , A = a ,D=2 (because consecutive odd)
9612 = 18/2(2a+(18-1)2)
9612 = 9(2a + 34)
9612/9 = 2a + 34
1068 = 2a + 34
1068 - 34 = 2a
1034 = 2a
517 = a !

I will go with A
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Bunuel
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

(A) 517
(B) 518
(C) 519
(D) 521
(E) 525


Note: we have total 18 consecutive integers. It means we got average by dividing the sum of 2 integers . Question is which 2 integers?


533 + 535 / 2 = 534.........................a bit common sense is required here.

If total number of integers were 19 , the average would be 10th integers. So, in this case we take care both 9th and 10th integer of the serious.


9th = 533

17-19-21-23-25-27-29-31


517 is the least among the odd integers.


The best answer is A.
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BuggerinOn
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525


\(\frac{x}{18}\)= \(534\)

x = 9,532

if each of the numbers were 501 than total would be 501*18 = 9,018

so if i deduct 1 from each of the odd numbers i get 534-18 = 516

so smallest odd number is 517
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BuggerinOn
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525

Since the average of 18 consecutive odd integers is 534, the average of the middle two integers is also 534. Therefore, the middle two integers are 533 and 535. There will be 8 odd integers less than 533 (and 8 more greater than 535). So the smallest of these integers is 533 - 8(2) = 533 - 16 = 517.

Alternate Solution:

If the smallest of these integers is x, then the largest of these integers will be x + 34. Since consecutive odd integers form an evenly spaced set of integers, the average is equal to the average of the smallest and the largest integers; thus:

(x + x + 34)/2 = 534

2x + 32 = 1068

2x = 1034

x = 517

Answer: A
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BuggerinOn
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525

mean falls in middle of sequence between between 9th and 10th terms
so, if least term is n,
then n+8*2+1=534→
n=517
A
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Given that the average(arithmetic mean) of 18 consecutive odd integers is 534 and we need to find the least of these integers

============================================================

Theory
    ‣‣‣ In Case of consecutive number with even number of term, Mean = Mean of Middle two terms

============================================================

Let the middle two terms are 2x-1 and 2x+1

=> Mean = mean of middle two terms = \(\frac{2x-1 + 2x+1}{2}\)= \(\frac{4x}{2}\) = 2x = 534

As there are 18 terms so the middle two terms will be \(9^{th}\) and \(10^{th}\) term
=> \(9^{th}\) term = 2x-1 = 534-1 = 533

First term or the lest term will be \(9^{th}\) term - 2*8 = 533 - 16 = 517

So, Answer will be A.
Hope it helps!

Watch the following video to Learn the Basics of Statistics

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