An arithmetic sequence of five terms is considered $good$ if it contains 19 and 20. For example, $18.5,19.0,19.5,20.0,20.5$ is a $good$ sequence. For every $good$ sequence, the sum of its terms is totalled. What is the total sum of all $good$ sequences?
Problem
Source: 2019 Malaysia OMK
Tags: arithmetic sequence, Arithmetic Progression, algebra
21.08.2019 15:30
hopeless404 wrote: An arithmetic sequence of five terms is considered $good$ if it contains 19 and 20. For example, $18.5,19.0,19.5,20.0,20.5$ is a $good$ sequence. For every $good$ sequence, the sum of its terms is totalled. What is the total sum of all $good$ sequences? Just count. Let $\Delta$ be the common progression of AP. $\Delta\in\left\{\pm1,\pm\frac 12,\pm\frac 13,\pm\frac 14\right\}$ 1) $\Delta=1$ implies 4 sequences : $19,20,21,22,23$ whose sum is $105$ $18,19,20,21,22$ whose sum is $100$ $17,18,19,20,21$ whose sum is $95$ $16,17,18,19,20$ whose sum is $90$ With total sum $390$ 2) $\Delta=\frac 12$ implies 3 sequences : $19,19.5,20,20.5,21$ whose sum is $100$ $18.5,19,19.5,20,20.5$ whose sum is $97,5$ $18,18.5,19,19.5,20$ whose sum is $95$ With total sum $292.5$ 3) $\Delta=\frac 13$ implies 2 sequences : $19,19\frac 13,19\frac 23,20,20\frac 13$ whose sum is $98\frac 13$ $18\frac 23,19,19\frac 13,19\frac 23,20$ whose sum is $96\frac 23$ With total sum $195$ 4) $\Delta=\frac 14$ implies 1 sequence : $19,19.25,19.5,19.75,20$ whose sum is $97.5$ With total sum $97.5$ And negative Delta give the same sequences in reverse order. Hence the result : $2\times(390+292.5+195+97.5)$ $=\boxed{1950}$
03.07.2024 11:56
why cant you have 20.19, 20.2, 20.21, 20.22
03.07.2024 12:29
hopeless404 wrote: An arithmetic sequence of five terms is considered $good$ if it contains 19 and 20. For example, $18.5,19.0,19.5,20.0,20.5$ is a $good$ sequence. For every $good$ sequence, the sum of its terms is totalled. What is the total sum of all $good$ sequences? Between 19 and 20 could be one up to four ds If it was 4, then both 19 and 20 are terminal terms If 3, it could be another term before 19 or another term after 20 in an increasing progression (and vice versa) and so on Answer = 5 × [ (19 + 20)/2 ] × (1 + 2 + 3 + 4) × 2 = 5 × 39 × 10 = 1950 We don’t mind the d itself because they all cancel each other out , for example between { 19-d , 19 , x , x , 20 } and { 19 , x , x , 20 , 20+d } , so the “5” is the number of terms, then then median, then the distinct progressions with positive ds, and double for d < 0
03.07.2024 13:43
Chanome wrote: why cant you have 20.19, 20.2, 20.21, 20.22 Because it must contain $19$ and $20$