Question

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Answer 1

We have the sequence 3, 6, 9, 12.

We can prove that this sequence is not geometric by calculating the ratios a2/a1 and compare it to a3/a2. If the ratios are not equal, the sequence is not geometric: there is no common ratio in the sequence.

`\begin{gathered} (a_2)/(a_1)=(6)/(3)=2 \\ (a_3)/(a_2)=(9)/(6)=1.5 \\ (a_2)/(a_1)\\eq(a_3)/(a_2)\Rightarrow\text{ not a geometric sequence} \end{gathered}`

We can prove it is a arithmetic sequence because it has a common difference d=3 between consecutive terms:

`\begin{gathered} a_2=a_1+3=3+3=6 \\ a_3=a_2+3=6+3=9 \end{gathered}`

Now, we have to calculate the sum.

We start by calculating a10, using the explicit function for an arithmetic sequence:

`a_(10)=f(10)=3+3\cdot(10-1)=3+3\cdot9=3+27=30`

Then, we apply the formula given for the sum of n terms:

`\begin{gathered} S=(n)/(2)(a_1+a_n) \\ S_(10)=(10)/(2)(3+30)=5\cdot33=165 \end{gathered}`

Answer:

The sequence is arithmetic.

The sum of the first 10 terms is S10=165.