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17 Jun 2019, 01:40
Kudos
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A
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Difficulty:
65%
(hard)
Question Stats:
66% (03:19) correct
34%
(03:13)
wrong
based on 242
sessions
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Not Attempted Yet
Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It continues in such a way that by adding 5 to the nth term, one obtains the (n + 2)th term. What is the sum of the first 19 members of this sequence?
A. 878
B. 900
C. 788
D. 928
E. 1022
A. 878
B. 900
C. 788
D. 928
E. 1022
17 Jun 2019, 02:04
Kudos
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Bunuel
Let first number = 23
Sequence is 23, 23+4, 23+5, 23+9, 23+10, . . . .
In first 19, there are 10 Odd terms & 9 Even terms
Odd terms \(t_1\), \(t_3\), \(t_5\). . . . . . . .\(t_{19}\) are 23, 23+5, 23+10 . . . . . . 23+45 --> AP series with common difference 5
Even terms \(t_2\), \(t_4\), \(t_6\). . . . . . . .\(t_{18}\) are 23+4, 23+9, 23+14, . . . . . . 23+44 --> AP series with common difference 5
Sum of Odd terms = n/2*(1st term + last term) = 10/2*(23 + 68) = 455
Sum of Even terms =9/2*(1st term + last term) = 9/2*(27 + 67) = 9/2*94 = 423
Total sum = 455 + 423 = 878
IMO Option A
Pls Hit Kudos if you like the solution
17 Jun 2019, 01:59
Kudos
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23,27,28,32,33,37..we can see that each pair sums up to be 50, 60, 70 and so on till 130 as there 9 pairs
Hence sums up to be 810 till 18th term
Since we know each pair has a difference of 4 therefore 17th and 18th term = 2x+4 =130
17 th term = 63
Hence 19th term = 63+5 =68
Hence sum = 810+68=878
Posted from my mobile device
Hence sums up to be 810 till 18th term
Since we know each pair has a difference of 4 therefore 17th and 18th term = 2x+4 =130
17 th term = 63
Hence 19th term = 63+5 =68
Hence sum = 810+68=878
Posted from my mobile device
