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1. part b i) i worked it out in my head, though not the most mathematical way, i got it t11+t14 = 50

but part b ii) is still confusing me, my teacher explained it to me and i still didnt get it, but it was the end of the school day so i couldnt stay. but anyway, i think theres more than one way to find the answer, so im open to other ways.
2. (Original post by ihatePE)

part b i) i worked it out in my head, though not the most mathematical way, i got it t11+t14 = 50

but part b ii) is still confusing me, my teacher explained it to me and i still didnt get it, but it was the end of the school day so i couldnt stay. but anyway, i think theres more than one way to find the answer, so im open to other ways.
For what it's worth, for part a)

Anytime it says write down, the answer is either obvious or a one liner, so you have to think it's related to 50 somehow.

to do it mathematically, note that t12 is a +(12-1)d and similarly t13 is...

Now t11 is .... and t14 is ....

Hence..

3. Nice question. Suggested approach:

a) write down an expression for t(12) + t(13) in terms of the first term and common difference of the series, using the standard formula for t(n) from the formula booklet;

b) write down an expression for the sum of the first 24 terms of an arithmetic series in terms of the first term and common difference of the series, using the standard formula for S(n) from the formula booklet;

c) look for the commonality between (a) and (b) and you're pretty well there.
4. when you add up an AP you are making equal pairs... the first + the last for instance. if there are 24 terms we can say that the 12th + the 13th terms will be one of the 12 equal pairs.
5. The algebraic solution to this is to use the results for Un and Sn for APs
Un = a + (n-1)d
Sn = (n/2)(a+L) = (n/2)(2a + N-1)d)
L = last term

(a)
50 = U12 + U13 = (a+11d)+(a+12d) = 2a+23d = 50
U11 + U14 = (a+10d) + (a+13d) = 2a +23d so also = 50

(b)
S24 = (24/2)(a + {a+23d}) = 12*(2a + 23d) = 600

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