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31 Dec 2018, 00:32
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Difficulty:
95%
(hard)
Question Stats:
50% (03:08) correct
50%
(02:50)
wrong
based on 127
sessions
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In an increasing sequence, the difference between any two successive numbers is twice that of the first term of the sequence. If S is a set of 3 successive numbers in this sequence and the largest number in the set S is 51 and the sum of all the numbers in S is 135. What is the 10th number in this sequence?
A. 39
B. 57
C. 66
D. 93
E. 145
A. 39
B. 57
C. 66
D. 93
E. 145
31 Dec 2018, 01:20
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Let the first term in the sequence be a
Given that the difference between any two successive numbers is twice that of the first term of the sequence. Hence, the common difference (d) = 2a
Let me the three consecutive integers in set S be the (n - 1)th, nth, and (n + 1)th terms
(n - 1)th term = a + (n - 2)2a
nth term = a + (n-1)2a
(n + 1)th term = a + (n)2a
Given that the largest term in S, i,e. the (n + 1)th term is 51. a + (n)2a = 51......equation 1
Again, the sum of all the numbers in S is 135, {a + (n - 2)2a} + {a + (n-1)2a} + {a + (n)2a} = 135.......equation 2
Solving equation 1 & equation 2, we get a = 3 & n = 8
The 10th number in this sequence = a + 9d = a + 18a = 19a (Since d=2a)
Plug in value of a, 19a = 57 (Since a = 3)
Hence the Correct Answer is Option B. 57
Given that the difference between any two successive numbers is twice that of the first term of the sequence. Hence, the common difference (d) = 2a
Let me the three consecutive integers in set S be the (n - 1)th, nth, and (n + 1)th terms
(n - 1)th term = a + (n - 2)2a
nth term = a + (n-1)2a
(n + 1)th term = a + (n)2a
Given that the largest term in S, i,e. the (n + 1)th term is 51. a + (n)2a = 51......equation 1
Again, the sum of all the numbers in S is 135, {a + (n - 2)2a} + {a + (n-1)2a} + {a + (n)2a} = 135.......equation 2
Solving equation 1 & equation 2, we get a = 3 & n = 8
The 10th number in this sequence = a + 9d = a + 18a = 19a (Since d=2a)
Plug in value of a, 19a = 57 (Since a = 3)
Hence the Correct Answer is Option B. 57
