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The sum of the first three terms of an arithmetic sequence is $24$ and the sum of the next three terms is $51$. Find the first term and the common difference.

Here's what I did:
working out

I listed the six terms below. The first three add up to $24$, but the next three don't add up to $51$. What am I doing wrong?

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  • $\begingroup$ You have taken the sum of the first six terms as 51, while it is the sum of terms 4,5 and 6. $\endgroup$ – GoodDeeds May 1 '16 at 9:12
  • $\begingroup$ So I'll subtract 6 from 4 and add 1? $\endgroup$ – AugieJavax98 May 1 '16 at 9:20
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According to given question:

Sum of first three terms is $24$, so:

$$a+(a+d)+(a+2d)=24$$

$$3a+3d=24$$

$$a+d=8\cdots(1)$$

And sum of next three terms is $51$, so: $$(a+3d)+(a+4d)+(a+5d)=51$$

$$3a+12d=51$$

$$a+4d=17\cdots(2)$$

On solving equestion $(1)$ and $(2)$, we get first term $a=5$ and common difference $d=3$.

AP series is $:5, 8, 11, 14, 17, 20, \cdots$

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A quick way to see the common difference $d$ mentally. The difference between term $n$ and term $n+3$ is $3d$. So the difference between those two sums gives you:

$3(3d) = 51 - 24$

$d = 3$

From that, you should be able to work out the first term very easily. Use the fact that the sum of the first three terms is $a + a + d + a + 2d = 3(a+d) = 24$, giving $a = 5$.

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  • $\begingroup$ Wow that's a cool short cut! $\endgroup$ – AugieJavax98 May 1 '16 at 9:32
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    $\begingroup$ @AugieJavax98 Thanks, I try to work out problems mentally at first. Helps to discover quick solutions. :) $\endgroup$ – Deepak May 1 '16 at 9:33
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With less algebra: The sum of the first $3$ terms of your sequence is $3$ times the average term, which is conveniently just the middle one. That is, the second term must be $\frac{24}{3} = 8$.

Similarly the fifth term (middle of the next three) is $\frac{51}{3}=17$.

In the three steps between the second and the fifth term, the increase has been $17-8=9$, so the step size is one-third of that, that is, $3$.

Since we already know the second term is $8$, the sequence must be $$5,8,11,14,17,20,\ldots $$

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