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?

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! Cheers for that, I knew my method was right!