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Quick Maths Q. (1 Viewer)

YonOra

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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

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u have to use simultaneous eqns.
The answer is 16.
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

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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

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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

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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

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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
1602296738571.png
 

devtrivedi

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

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YonOra

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u have to use simultaneous eqns.
The answer is 16.
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.
 

YonOra

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A faster method:



The formula here is derived by the above method as:

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

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Short-cutting your method:

Having found equations (1) and (2), rather than solving for and , perform the subtraction (1) - (2) to get



and then recognise that



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

YonOra

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Short-cutting your method:

Having found equations (1) and (2), rather than solving for and , perform the subtraction (1) - (2) to get



and then recognise that



and so you have the answer.
Smart
 

CM_Tutor

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Thanks, YonOra.

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

By this, I mean: keep your eye on your goal.

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

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