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:
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?
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 $:5, 8, 11, 14, 17, 20, \cdots$
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 $a = 5$.
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 $$5,8,11,14,17,20,\ldots $$
