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:

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

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

find a_{100}

Thank you.

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:

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

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

find a_{100}

Thank you.

Please edit your post.

The formatting makes it impossible to know what you mean.

Re: Finding 'a' given two summations

Given:

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

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

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.

Re: Finding 'a' given two summations

Quote:

Originally Posted by

**Kevmck** Given:

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

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

find $\displaystyle a_{100}$

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

$\displaystyle \sum\limits_{j = 0}^{98} {\left( {a_{j + 1} + 5} \right)} = \sum\limits_{j = 1}^{99} {a_j } + 5(99)$

Re: Finding 'a' given two summations

I'm not sure I really understand, but thank you for the help.

Re: Finding 'a' given two summations

Hello, Kevmck!

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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

Subtract: .$\displaystyle a_{100} \:=\:150$

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.