Register  
 
 About Us | Help | Sign in
 
HomeForumsArticlesRevision NotesPersonal StatementsUniversity GuidesHealth & RelationshipsCareersFreshers 2008
Create New Article | Editing Help  
   

Advanced Search
STEP I 1990 question 4 solution

From The Student Room

TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1990 question 4 solution

Arithmetic Progressions

If, \displaystyle a_{1}, a_{2}, \; ... \;, a_{n} is an arithmetic progression, then

\displaystyle S_{n} = \frac{n}{2}(2a + (n-1)d) = \frac{n}{2}(2a_{1} + (n-1)(a_{2}-a_{1})) = \frac{n}{2}((3-n)a_{1} + (n-1)a_{2})


Double Arithmetic Progressions

If \displaystyle b_{2}-b_{1}, \; b_{3}-b_{3}, \; ... \;, b_{n+1}-b_{n}, \; ... is an arithmetic progression, then

\displaystyle b_{n} = a + (n-2)d + b_{n-1}

\displaystyle \Rightarrow b_{n} = \left( \sum^{n}_{r=2} a + (r-2)d\right) + b_{1} = (n-1)a + d \left( \sum^{n}_{r=2}r-2 \right) + b_{1}

\displaystyle \sum^{n}_{r=2}(r-2) = \sum^{n-2}_{k=1} k = \frac{1}{2}(n-1)(n-2)

\displaystyle \Rightarrow b_{n} = (n-1)a + \frac{1}{2}(n-1)(n-2)d + b_{1}

\displaystyle \sum_{r=1}^{n} b_{r} = a\left(\sum_{r=1}^{n} n-1\right) + \frac{d}{2}\left(\sum_{r=1}^{n} n^{2} - 3n + 2 \right) + nb_{1}

\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{2}n(n-1)a + \frac{1}{6}n(n-1)(n-2)d + nb_{1}

\displaystyle a = b_{2} - b_{1}, \; d = (b_{3} - b_{2}) - (b_{2} - b_{1}) = b_{3} - 2b_{2} + b_{1}

Several minutes of tedious algebra later:

\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{6}n\left((n^{2} - 6n + 11)b_{1} - (n-1)(2n-7)b_{2} + (n-1)(n-2)b_{3}\right)


Factorial Progressions

b_{4}-b_{2} = d, \; b_{6}-b_{4} = 2d, \; 220 - b_{6} = 6d

b_{2} - 1 = a, \; b_{n} - b_{n-1} = a + (n-2)x

(b_{3} - b_{2}) + (b_{4} - b_{3}) = b_{4} - b_{2} = 2a + 3x = d

(b_{5}-b_{4}) + (b_{6} - b_{5}) = b_{6} - b_{4} = 2a + 7x = 2d

2a + 7x = 4a + 6x \Rightarrow x = 2a, \; d = 8a

(b_{2}-1) + (b_{3}-b_{2}) + \cdots + (b_{6}-b_{5}) = b_{6} - 1 = 5a + 10x

b_{6} = 220 - 6d = 5a + 10x + 1 \Rightarrow 5a + 10x + 6d = 219

5a + 20a + 48a = 219 \Rightarrow 73a = 219

\Rightarrow a = 3, \; x = 6, \; d = 24

Putting a = 3, \; d = 6, \; b_{1} = 1 into

\displaystyle \sum_{r=1}^{n} b_{r} = \frac{1}{2}n(n-1)a + \frac{1}{6}n(n-1)(n-2)d + nb_{1}

we get:

\displaystyle S_{n} = \frac{3}{2}n(n-1) + n(n-1)(n-2) + n = \frac{1}{2}n(2n^{2} - 3n + 3)

Solution by Dystopia.

Retrieved from "http://thestudentroom.co.uk/wiki/STEP_I_1990_question_4_solution"
Article Tools:  Create New Article  |  Report This Article  |  This Article's History  | 
collapse
Recent Threads
 
collapse Cyrptic football club names quiz
started by: Fusion
forum: Football
replies: 16
last post: 1 Minute Ago
collapse my 21 year old friend just got cancer, how can i be supportive?
started by: Anonymous
replies: 1
last post: 1 Minute Ago
collapse I like him. He has a girlfriend but i think he likes me.
started by: Anonymous
replies: 16
last post: 1 Minute Ago
collapse Graph of (4-x)/(x-1)
started by: ST18
forum: Maths
replies: 11
last post: 1 Minute Ago
collapse Oxford application form-confused!!!Please help!
started by: eve_22
replies: 4
last post: 1 Minute Ago