# Quick Maths Q. (1 Viewer)

#### YonOra

##### Well-Known Member
Hello, hello!

I'll cut right to it.

I'm confused why the answer is 16 - and feel especially stupid because you get that from 249 - 233, however i don't see the connection. Can someone explain this to me?

The sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. Find the 50th term of the series.

Cheers

#### devtrivedi

##### New Member
u have to use simultaneous eqns.
Your difference should be 551/1225 and the first term should be -151/25
Plug that into the formula for the nth term to get the answer

#### Pikapizza

##### Member
The first statement that “the sum of 50 terms of an arithmetic series is 249” basically means that a1 + a2 + ...... + a50 = 249 (just think a1, a2... a50 as numbers in order in an arithmetic series).
The second statement: “sum of 49 terms of the series is 233” can be written as a1 + a2 + .... + a49 = 233.
Therefore to find the 50th term (a50) you subtract both sums to find a50.
I think the main reason you didn’t get the connection was that the question did not specify where the sum starts (they should have specified the sum of the first 50/49 numbers of the same series, which might have helped with you getting it).

#### YonOra

##### Well-Known Member
The first statement that “the sum of 50 terms of an arithmetic series is 249” basically means that a1 + a2 + ...... + a50 = 249 (just think a1, a2... a50 as numbers in order in an arithmetic series).
The second statement: “sum of 49 terms of the series is 233” can be written as a1 + a2 + .... + a49 = 233.
Therefore to find the 50th term (a50) you subtract both sums to find a50.
I think the main reason you didn’t get the connection was that the question did not specify where the sum starts (they should have specified the sum of the first 50/49 numbers of the same series, which might have helped with you getting it).
Thank you! And yeah, I understand what you've done. I just don't understand this
Therefore to find the 50th term (a50) you subtract both sums to find a50.

#### devtrivedi

##### New Member
The first statement that “the sum of 50 terms of an arithmetic series is 249” basically means that a1 + a2 + ...... + a50 = 249 (just think a1, a2... a50 as numbers in order in an arithmetic series).
The second statement: “sum of 49 terms of the series is 233” can be written as a1 + a2 + .... + a49 = 233.
Therefore to find the 50th term (a50) you subtract both sums to find a50.
I think the main reason you didn’t get the connection was that the question did not specify where the sum starts (they should have specified the sum of the first 50/49 numbers of the same series, which might have helped with you getting it).
good method

#### Pikapizza

##### Member
Thank you! And yeah, I understand what you've done. I just don't understand this
Therefore to find the 50th term (a50) you subtract both sums to find a50.
Excuse my terrible handwriting: but this hopefully should explain why you subtract the two sums

#### devtrivedi

##### New Member
This is how I did it. But Pikapizza's method is better.

#### Attachments

• 426 KB Views: 14

#### YonOra

##### Well-Known Member
u have to use simultaneous eqns.
Your difference should be 551/1225 and the first term should be -151/25
Plug that into the formula for the nth term to get the answer
Ye very true, and i did this - but it's too messy and very vulnerable to dumb mistakes.

#### devtrivedi

##### New Member
Ye very true, and i did this - but it's too messy and very vulnerable to dumb mistakes.
right

#### YonOra

##### Well-Known Member
Excuse my terrible handwriting: but this hopefully should explain why you subtract the two sums
View attachment 29131
Ye talk about visual learning haha. Thanks, really appreciate it.

#### CM_Tutor

##### Moderator
Moderator
A faster method:

\bg_white \begin{align*} S_n &= S_{n-1} + T_n \\ T_n &= S_n - S_{n-1} \\ &= 249 - 233 \\ &= 16 \end{align*}

The formula here is derived by the above method as:

\bg_white \begin{align*} T_n &= S_n - S_{n-1} \\ &= \left(T_1 + T_2 + T_3 + ... + T _{50}\right) - \left(T_1 + T_2 + T_3 + ... + T _{49}\right) \\ &= T_{50} \end{align*}

#### CM_Tutor

##### Moderator
Moderator
This is how I did it. But Pikapizza's method is better.

Having found equations (1) and (2), rather than solving for $\bg_white a$ and $\bg_white d$, perform the subtraction (1) - (2) to get

$\bg_white a + 49d = 16$

and then recognise that

$\bg_white T_{50} = a + (50 - 1)d = a + 49d$

and so you have the answer.

#### YonOra

##### Well-Known Member
A faster method:

\bg_white \begin{align*} S_n &= S_{n-1} + T_n \\ T_n &= S_n - S_{n-1} \\ &= 249 - 233 \\ &= 16 \end{align*}

The formula here is derived by the above method as:

\bg_white \begin{align*} T_n &= S_n - S_{n-1} \\ &= \left(T_1 + T_2 + T_3 + ... + T _{50}\right) - \left(T_1 + T_2 + T_3 + ... + T _{49}\right) \\ &= T_{50} \end{align*}
woa this is cool. I've never seen this (I literally just started this topic today - in Year 11, albeit only for another 3 days).

#### devtrivedi

##### New Member

Having found equations (1) and (2), rather than solving for $\bg_white a$ and $\bg_white d$, perform the subtraction (1) - (2) to get

$\bg_white a + 49d = 16$

and then recognise that

$\bg_white T_{50} = a + (50 - 1)d = a + 49d$

and so you have the answer.
oh true, i didnt see that

#### YonOra

##### Well-Known Member

Having found equations (1) and (2), rather than solving for $\bg_white a$ and $\bg_white d$, perform the subtraction (1) - (2) to get

$\bg_white a + 49d = 16$

and then recognise that

$\bg_white T_{50} = a + (50 - 1)d = a + 49d$

and so you have the answer.
Smart

#### CM_Tutor

##### Moderator
Moderator
Thanks, YonOra.

Regarding this short cut, this is an example of a valuable exam strategy that is worth bearing in mind.

In this case, you formed two simultaneous equations in two unknowns and you did the obvious thing - solved them to find $\bg_white a$ and $\bg_white d$ - but you didn't stop long enough to recognise that you weren't actually seeking $\bg_white a$ or $\bg_white d$. Instead, you were seeking $\bg_white T_{50}$ which happens to be $\bg_white a + 49d$. My short-cut simply recognises that the goal is this latter expression and uses the simultaneous equations to reach it.