Arithmetic Pdf
Need help with Maths A level question!!?
Hi i need help with this question:
The tenth term of an arithmetic progression is equal to twice the fourth term. the twentieth term of the progression is 44.
find the first term and the common difference. find the sum of the first 50 terms.
it is in a past OCR paper:
http://www.ocr.org.uk/download/pp_09_jun/ocr_40110_pp_09_jun_gce_4722_01.pdf
but there isnt a mark scheme for it and i would really like to know the answer and how people worked it out.
Thanks
Matt
Wolfe, if 44 is meant to be the 20th term then ur method doesnt add up:
1st term = 6.6
2nd = 8.8
3rd = 11
4th = 13.2
5th = 15.4
6th = 17.6
7th = 19.8
8th = 22
9th = 24.2
10th = 26.4
11th= 28.6
12th = 30.8
13th = 33
14th = 35.2
15th = 37.4
16th = 39.6
17th = 41.8
18th = 44
so with ur method the 18th term is 44, not the 20th
Let the first term be a and the common difference d.
The nth term is a + (n – 1)d.
The 10th term is a + 9d.
The 4th term is a + 3d.
Hence a + 9d = 2(a + 3d) = 2a + 6d.
Rearranging, we find that a = 3d. (1)
The 20th term is a + 19d.
So a + 19d = 44.
But from (1), a = 3d, so 3d + 19d = 44.
Hence 22d = 44 and d = 2.
Hence a = 3×2 = 6.
The first term is 6 and the common difference is 2.
The sum to n terms is ½n(2a + (n – 1)d).
Setting n = 50, the sum to 50 terms is 25(12 + 49×2) = 2750.
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