SOLUTION: The sum of the first 12 terms of an AP (Arithmetic Progression) is 222, the sum of the first 5 terms is 40. Write out the first four terms of the series.

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Question 142024: The sum of the first 12 terms of an AP (Arithmetic Progression) is 222, the sum of the first 5 terms is 40. Write out the first four terms of the series.
Answer by nabla(475) About Me  (Show Source):
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The sum of the first n terms of an arithmetic progression is na%2B%28n%28n-1%29r%29%2F2.
Using this formula with n=12 and n=5,
222=12a%2B%2812%2812-1%29r%29%2F2
40=5a%2B%285%285-1%29r%29%2F2
simplifying a bit:
222=12a%2B66r
40=5a%2B10r
and simplifying a bit more:
37=2a%2B11r
8=a%2B2r
We now solve this system of equations.
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++2%5Ca+%2B+11%5Cr+=+37%2C%0D%0A++++1%5Ca+%2B+2%5Cr+=+8+%29%0D%0A++We'll use substitution. After moving 11*r to the right, we get:
2%2Aa+=+37+-+11%2Ar, or a+=+37%2F2+-+11%2Ar%2F2. Substitute that
into another equation:
1%2A%2837%2F2+-+11%2Ar%2F2%29+%2B+2%5Cr+=+8 and simplify: So, we know that r=3. Since a+=+37%2F2+-+11%2Ar%2F2, a=2.

Answer: system%28+a=2%2C+r=3+%29.


a=2, r=3.
The first 4 terms of the series would be, then:
2,(2+3),(2+2(3)),(2+3(3)) which is
2,5,8,11
In order to check, which is always a good idea, we can only iterate:
2+5+8+11+14=40 (first part is correct)
2+5+8+11+14+17+20+23+26+29+32+35=222 (second part is correct).