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09 Mar 2016, 13:31
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B
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Difficulty:
85%
(hard)
Question Stats:
43% (01:48) correct
57%
(01:57)
wrong
based on 235
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What is the 6th term of the Arithmetic sequence?
(1) The sum of the 6th to the 12th term of the sequence is 77.
(2) The sum of the 2nd to the 10th term of the sequence is 108.
(1) The sum of the 6th to the 12th term of the sequence is 77.
(2) The sum of the 2nd to the 10th term of the sequence is 108.
09 Mar 2016, 20:36
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We need to find T6 = a+5d
(1) The sum of the 6th to the 12th term of the sequence is 77.
T6 + T7 + T8 + T9 + T10 + T11 + T12 = 77
Sum to 12th term - Sum to 5th term = 77
=>6[ 2a + 11d] - 5/2[2a+4d] = 77
=> 12a +66d -5a - 10d = 77
=> 7a + 56d = 77
=> a + 8d = 11
Not sufficient as we have 2 variables and 1 equation
(2) The sum of the 2nd to the 10th term of the sequence is 108
T2+ T3 + T4 + T5 + T6 + T7 +T8 + T9 + T10 = 108
In the above AP , T6 will be the mean of the sequence.
9*T6 = 108
=>T6 = 12
Sufficient
Answer B
(1) The sum of the 6th to the 12th term of the sequence is 77.
T6 + T7 + T8 + T9 + T10 + T11 + T12 = 77
Sum to 12th term - Sum to 5th term = 77
=>6[ 2a + 11d] - 5/2[2a+4d] = 77
=> 12a +66d -5a - 10d = 77
=> 7a + 56d = 77
=> a + 8d = 11
Not sufficient as we have 2 variables and 1 equation
(2) The sum of the 2nd to the 10th term of the sequence is 108
T2+ T3 + T4 + T5 + T6 + T7 +T8 + T9 + T10 = 108
In the above AP , T6 will be the mean of the sequence.
9*T6 = 108
=>T6 = 12
Sufficient
Answer B
10 Mar 2016, 02:23
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Bunuel
Arithmetic Sequence: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is constant.
Ex: a, a + d, a + 2d. or 2, 6, 10, 14.
Required: 6th term of the Arithmetic sequence
Or a + 5d = ?
Statement 1: The sum of the 6th to the 12th term of the sequence is 77
If we assume the first term to be a
Then 6th term = a + 5d, 7th term= a + 6d, 8th term = a + 7d, 9th term = a + 8d, 10th term = a + 9d, 11th term = a + 10d and 12th term = a + 11d
Sum = 7a + 66d = 77
We cannot find a + 5d from here.
INSUFFICIENT
Statement 2: The sum of the 2nd to the 10th term of the sequence is 108
2nd term = a + d, 3rd term = a +2d, 4th term = a + 3d, 5th term = a + 4d, 6th term = a + 5d, 7th term = a + 6d, 8th term = a + 7d, 9th term = a + 8d, 10th term = a + 9d
Sum = 9a + 45d = 108
Or a + 5d = 12
Hence 6th term = 12
SUFFICIENT
Option B
