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?
Best Answer
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 $:5, 8, 11, 14, 17, 20, \cdots$