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15 May 2020, 01:15
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A
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Difficulty:
35%
(medium)
Question Stats:
75% (02:34) correct
25%
(02:51)
wrong
based on 79
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In a sequence of numbers in which the difference of any two successive terms is a constant, the second term is 8 and the eighths term is 2 more than thrice the second term. What is the sum of the first eight terms of this sequence?
A. 124
B. 110
C. 108
D. 106
E. 96
Are You Up For the Challenge: 700 Level Questions
A. 124
B. 110
C. 108
D. 106
E. 96
Are You Up For the Challenge: 700 Level Questions
15 May 2020, 01:43
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Bunuel
If the first term = a
COmmon difference = d
Then, Second Term = a+d = 8
Then, Eigth Term = a+7d = 3*(a+d)+2
i.e. a+7d = 3*(a+d)+2
i.e. 2a+2 = 4d
i.e. a +1 = 2d i.e .a = 2d - 1
Also, a+d = 8
i.e. (2d-1)+d = 8
i.e. d = 3
and a = 5
First 8 terms are {5, 8, 11, 14, 17, 20, 23, 26}
Sum of terms in A.P. = (n/2)[First term+Last Term] = (8/2)*(5+26) = 4*31 = 124
Answer: Option A
15 May 2020, 02:36
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Bunuel
Solution
- • Since the difference of any two successive terms is a constant, so the given sequence is in Arithmetic progression.
• Now, let us assume that the first term of the A.P. is a and the common difference is d.
• Second term \(= a + d = 8 …………Eq.(i)\)
• Eighth term \(= a + 7d = 3*8 + 2 = 26 ….Eq.(ii)\)
• Subtracting second term from eighth term(or, Eq.(i) from Eq.(ii)) , we get,
- o \(6d = 18 ⟹ d = 3\)
o Thus, \(a = 8 – 3 = 5\)
