altairahmad wrote:
Bunuel EMPOWERgmatRichC VeritasKarishma chetan2uWhy can't we solve this using AP formulae. I am consistently getting (D) as the answer. Where am I going wrong ?
\(a_{11} (avg) = a_1 + 10d\)
\(n = 4(a_{11}) = 4(a_1 + 10d)\)
\(S (sum) = \frac{21}{2}(2a_1 + 20d)\)
Putting in any dummy values in resulting expression for \(\frac{n}{S}\) will give (D) as the fraction.
Please help.
You are not given that the numbers in the list are in arithmetic progression. You are only given that n is 4 times the avg of other 20 numbers.
The numbers could be 1, 5, 6, 6, 6, 6, ... 6, 12
with an average of 6. Then n would be 24.
1, 5, 6, 6, 6, 6, 6, 6, .. 6, 12, 24 is not an AP.
Also, you are given that n = 4 * avg of other 20 numbers.
Even if this were an AP in which n were the last term, a11 is not the avg of first 20 numbers. The avg of first 20 numbers would be avg of 10th and 11th numbers.
_________________
Karishma
Owner of Angles and Arguments at https://anglesandarguments.com/
NOW PUBLISHED - READING COMPREHENSION MODULE
For Individual GMAT Study Modules, check Study Modules >
For Private Tutoring, check Private Tutoring >