# Finding 'a' given two summations

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• Nov 18th 2012, 02:55 PM
Kevmck
Finding 'a' given two summations
I've been stuck on how to do this question for over an hour. I clearly don't understand how to do it, hopefully somebody can help explain this to me?

Given:

$\sum\limits_{k = 1}^{100}{(6a_k + 8)} = 2000$

$\sum\limits_{j = 0}^{98}{(a_j_+_1)} = 545$

find a100

Thank you.
• Nov 18th 2012, 03:07 PM
Plato
Re: Finding 'a' given two summations
Quote:

Originally Posted by Kevmck
I've been stuck on how to do this question for over an hour. I clearly don't understand how to do it, hopefully somebody can help explain this to me?

Given:

$\sum\limits_{k = 1}^{100}{(6ak + 8)} = 2000$

$\sum\limits_{j = 0}^{98}{(aj+1​)} = 545$

find a100

Thank you.

Please edit your post.
The formatting makes it impossible to know what you mean.
• Nov 18th 2012, 03:12 PM
Kevmck
Re: Finding 'a' given two summations
Given:

$\sum\limits_{k = 1}^{100}{(6a_k + 8)} = 2000$

$\sum\limits_{j = 0}^{98}{(a_j_+_1 + 5)} = 545$

find a100​

Sorry, I wasn't sure how to fix the formatting before, I shouldn't have left it like that until I figured out how. Fixed.
• Nov 18th 2012, 03:25 PM
Plato
Re: Finding 'a' given two summations
Quote:

Originally Posted by Kevmck
Given:

$\sum\limits_{k = 1}^{100}{(6a_k + 8)} = 2000$

$\sum\limits_{j = 0}^{98}{(a_j_+_1 + 5)} = 545$

find $a_{100}​$

$\sum\limits_{k = 1}^{100} {\left( {6a_k + 8} \right)} = 6\sum\limits_{k = 1}^{100} {a_k } + 8(100)$

$\sum\limits_{j = 0}^{98} {\left( {a_{j + 1} + 5} \right)} = \sum\limits_{j = 1}^{99} {a_j } + 5(99)$
• Nov 18th 2012, 03:42 PM
Kevmck
Re: Finding 'a' given two summations
I'm not sure I really understand, but thank you for the help.
• Nov 18th 2012, 04:41 PM
Soroban
Re: Finding 'a' given two summations
Hello, Kevmck!

Quote:

$\text{Given: }\:(1)\;\sum\limits_{k = 1}^{100}{(6a_k + 8)} \:=\: 2000 \qquad (2)\;\sum\limits_{j = 0}^{98}{(a_j_+_1 + 5)} = 545$

$\text{Find }a_{100}.$

$(1)\;\sum^{100}_{k=1}(6a_k + 8) \;=\;2000$

. $6\sum^{100}_{k=1} a_k + \sum^{100}_{k=1}8 \;=\;2000$

. . $6\sum^{100}_{k=1}a_k + 800 \;=\;2000$

. . . . . . $6\sum^{100}_{k=1}a_k \;=\;1200$

. . . . . . . $\sum^{100}_{k=1}a_k \;=\;200\;\;[3]$

$(2) \;\sum^{98}_{j=0}\left(a_{j+1} + 5\right) \;=\;545$

. $\sum^{98}_{j=0}a_{j+1} + \sum^{98}_{j=0}5 \;=\;545$

. . $\sum^{98}_{j=0}a_{j+1} + 495 \;=\;545$

. . . . . . $\sum^{98}_{j=0}a_{j+1} \;=\;50\;\;[4]$

From [3], we have: . $a_1 + a_2 + a_3 + \cdots + a_{99} + a_{100} \:=\:200$

From [4], we have: . $a_1 + a_2 + a_3 + \cdots + a_{99} \qquad\quad =\;\;50$

Subtract: . $a_{100} \:=\:150$
• Nov 19th 2012, 11:50 AM
Kevmck
Re: Finding 'a' given two summations
Thank you Soroban. In writing out potential solutions after getting the help from Plato, I eventually came up with 200 and 50 but I didn't know what I was supposed to do with them from there. This example has helped solidify what was happening and what the question was asking for. Thank you again.