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30 Nov 2021, 07:00
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Difficulty:
25%
(medium)
Question Stats:
86% (02:17) correct
14%
(02:05)
wrong
based on 57
sessions
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In a certain sequence, the first term is 1 and each term after the first is 20 times the previous term. If the sum of the 20th and 21st terms is divided by the difference between the 16th and 15th terms, the result is approximately equal to
A) \(20^4\)
B) \(20^5\)
C) \(20^6\)
D) \(20^7\)
E) \(20^8\)
A) \(20^4\)
B) \(20^5\)
C) \(20^6\)
D) \(20^7\)
E) \(20^8\)
30 Nov 2021, 22:25
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This is a textbook Geometric Progression question.
The general form of a term in a GP is \(ar^{n-1}\)
where a is the first term, r is the common ratio and n is the number of terms.
Here, the first term is given as 1, so a = 1.
Since each number in the sequence is the previous term multiplied by 20, r = 20.
We will first list down all the terms that is needed in this question.
20th term = \(ar^{n-1}\) = \(1*20^{19}\)
21th term = \(ar^{n-1}\) = \(1*20^{20}\)
16th term = \(ar^{n-1}\) = \(1*20^{15}\)
15th term = \(ar^{n-1}\) = \(1*20^{14}\)
sum of the 20th and 21st terms = \(20^{19} + 20^{20}\)
difference between the 16th and 15th terms = \(20^{15} - 20^{14}\)
sum of the 20th and 21st terms is divided by the difference between the 16th and 15th terms = \(\frac{20^{19} + 20^{20}}{20^{15} - 20^{14}}\)
\(\frac{20^{19}(1 + 20)}{20^{14}(1 - 20)}\)
\(\frac{20^{5}(1 + 20)}{(1 - 20)}\)
Since the question asks for approximately and the value of 21/19 is equal to around 1.105 it can be negligible.
Hence, \(20^5\) Option B is the answer to this question.
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The general form of a term in a GP is \(ar^{n-1}\)
where a is the first term, r is the common ratio and n is the number of terms.
Here, the first term is given as 1, so a = 1.
Since each number in the sequence is the previous term multiplied by 20, r = 20.
We will first list down all the terms that is needed in this question.
20th term = \(ar^{n-1}\) = \(1*20^{19}\)
21th term = \(ar^{n-1}\) = \(1*20^{20}\)
16th term = \(ar^{n-1}\) = \(1*20^{15}\)
15th term = \(ar^{n-1}\) = \(1*20^{14}\)
sum of the 20th and 21st terms = \(20^{19} + 20^{20}\)
difference between the 16th and 15th terms = \(20^{15} - 20^{14}\)
sum of the 20th and 21st terms is divided by the difference between the 16th and 15th terms = \(\frac{20^{19} + 20^{20}}{20^{15} - 20^{14}}\)
\(\frac{20^{19}(1 + 20)}{20^{14}(1 - 20)}\)
\(\frac{20^{5}(1 + 20)}{(1 - 20)}\)
Since the question asks for approximately and the value of 21/19 is equal to around 1.105 it can be negligible.
Hence, \(20^5\) Option B is the answer to this question.
Click here for more GMAT questions with Video solutions
