# Maths help please: Arithmetic Progression question

Discussion in 'General Discussion' started by Mad old tory, 10 May 2006.

# Location: Limehouse

Hi guys, got a question here that i can't seem to make sense

The sum of the first 12 terms of an AP is 222, the sum of the first 5 terms is 40. Write out the first four terms of the series.

Right I'm using te Sn=n/2[2a+(n-1)d] equation to form a simultaneous equation to work out a and d:

222=6[2a+11d]
40=5/2[2a+4d]

Multiplying the brackets leaves me with:

222=12a+66d
40=5a+10d

Multiplying the first one by 5 and the second by 12 to get a common a of 60 lets me find d, but it doesn't give me the right answer. for the record, a is 2 and d is 3, but i can't get those answers. I know I've obviously done something silly, or might even be doing the completely wrong thing, can someone point me in the right direction please?

2. Dave

# Location: Sacramento CA

222=12a+66d
40=5a+10d

Is correct - using row-reduced echelon form on my TI83 gives a=2, d=3

Long way:

1110 = 60a + 330d (top one times 5)
480 = 60a + 120d (bottom one times 12)

Subtract bottom from top:
630 = 210d

Thus d = 3

Substitution:
40 = 5a +10(3)

Thus a = 2

Think you made some basic mistake you'll kick yourself for yeh i have done, had 330-120 down as 110 instead of 210, can you tell I've been revising all day!   