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Three consecutive terms in an arithmetic sequence are $3e^{p}$, $5$, $3e^{r}$ - AQA - A-Level Maths Mechanics - Question 9 - 2017 - Paper 2
Question 9
Three consecutive terms in an arithmetic sequence are $3e^{p}$, $5$, $3e^{r}$. Find the possible values of $p$. Give your answers in an exact form. --- Prove that... show full transcript
Worked Solution & Example Answer:Three consecutive terms in an arithmetic sequence are $3e^{p}$, $5$, $3e^{r}$ - AQA - A-Level Maths Mechanics - Question 9 - 2017 - Paper 2
Step 1
Find the possible values of $p$
Answer
To determine the values of , we first recognize that in an arithmetic sequence, the difference between consecutive terms must be constant.
Setting up the equation, we have: This simplifies to:
Next, we can express it in terms of the common difference:
Equating the differences gives: This can be rearranged to give us:
Now, substituting in terms of leads us to a quadratic equation:
ightarrow r = p + 1$$ Substituting back, we find: $$3e^{2p+1} + 3e^{p} - 10 = 0$$ Solving this quadratic equation using the quadratic formula yields: $$p = - rac{1}{2} ext{ and } p = -1$$ So the possible values of $p$ are $p = - rac{1}{2}, -1$.Step 2
Prove that there is no possible value of $q$
Answer
To prove this, we start by assuming that , , and are consecutive terms of a geometric sequence.
In a geometric sequence, the ratio between consecutive terms must be the same. We can therefore write: rac{5}{3e^{-q}} = rac{3e^{q}}{5}
Cross-multiplying gives: Which simplifies to:
This is a contradiction as 25 does not equal 9. Therefore, we conclude that there is no possible value of for which , , can be consecutive terms of a geometric sequence.