1. ## Arithmetic Progression question

Hey guys, I'm having problems with this question:

The sum of the first 100 terms in an arithmetic progression with first term a and common difference d is T. The sum of the first 50 even-numbered terms, i.e. 2nd, 4th, 6th, ..., 100th, is 0.5T + 100. Find the value of a in terms of T. (Answer: a = T/100 - 198)

Thanks in advance if you could help me with the above question!

2. Originally Posted by margaritas
Hey guys, I'm having problems with this question:

The sum of the first 100 terms in an arithmetic progression with first term a and common difference d is T. The sum of the first 50 even-numbered terms, i.e. 2nd, 4th, 6th, ..., 100th, is 0.5T + 100. Find the value of a in terms of T. (Answer: a = T/100 - 198)

Thanks in advance if you could help me with the above question!
You have a sum,
$a+(a+d)+(a+2d)+...+(a+99d)$
This is what I like to do since I hate formulas because it is better to understand than to remember the facts. So I am not going to use the sum formula (in fact I do not even know it!) instead rewrite this like this.
$(\underbrace{a+a+...+a}_{100 \mbox{ times}})+d(1+2+...+99)$
You should know that the sum of the first "n" integers is:
$\frac{n(n+1)}{2}$
Thus,
$100a+d\left( \frac{100(99)}{2} \right)$
Thus,
$T=100a+4950d$ (1)
---
The second sum is,
$(a+d)+(a+3d)+...+(a+99d)$
I think it is easier to express it in "sigma notation".
$\sum_{k=1}^{50}a+(2k-1)d$
Thus,
$\sum_{k=1}^{50}a+2kd-d$
Use the fact, the sum of the first "n" integers,
In sigma notation it says,
$\sum_{k=1}^n k=\frac{n(n+1)}{2}$
Thus, in this case,
$50a+2\left( \frac{50(51)}{2} \right) d - 50 d$
Note the 50 in front on the "a" and "d" because you add them 50 times.
$50a+2500d=\frac{1}{2}T+100$ (2)
Now you have two equation (1) and (2) you should be able to solve for "a" in terms of "T".

3. Thanks ThePerfectHacker for your solution but I still dont quite get it.

EDIT: Got the answer, finally! Thanks once again!

RE-EDIT: No actually, what if I would to use formulas to solve this question? How would the solution be presented?

4. Originally Posted by margaritas
RE-EDIT: No actually, what if I would to use formulas to solve this question? How would the solution be presented?
The last two equations that you solve are identical.

5. Hello, margaritas!

I had to baby-talk my way through this one.
See if you like my approach . . .

The sum of the first 100 terms in an A.P. with first term $a$ and common difference $d$ is $T.$
The sum of the first 50 even-numbered terms is $\frac{1}{2}T + 100$
Find the value of $a$ in terms of $T.$ .(Answer: $a \:= \:\frac{T}{100} - 198$)

We're expected to know the formula for the sum of the first $n$terms of an A.P.
. . . $S_n\;=\;\frac{n}{2}\left[2a + d(n-1)\right]$

The sum of the first 100 terms is $T.$
. . $S_{100} \:=\:\frac{100}{2}\left[2a + 99d\right] \:=\:T$

. . which simplifies to: . $100a + 4950d\:=\:T$ [1]

The first fifty even terms begin with $a_2 = a + d$ and have a common difference of $2d.$

Hence: . $S_{\text{50 even}} \:=\:\frac{50}{2}\left[2(a+d) + 49(2d)\right] \:=\:\frac{1}{2}T + 100$

. . which simplifies to: . $100a + 5000d\:=\:T + 200$ [2]

Subtract [1] from [2]: . $50d = 200\quad\Rightarrow\quad d = 4$

Substitute into [1]: . $100a + 4950(4)\:=\:T\quad\Rightarrow\quad 100a = T - 19800$

Therefore: . $\boxed{a\:=\:\frac{T}{100} - 198}$

6. Yay thanks so much Soroban, now I know how to get the second equation