Question

Find the $6^{th}$ term of the sequence $\dfrac{7}{3},\dfrac{35}{6},\dfrac{121}{12},\dfrac{335}{24}....$
1. $\dfrac{2113}{96}$
2. $\dfrac{2112}{96}$
3. $\dfrac{2111}{96}$
4. $\dfrac{865}{48}$

Answer 1

To get what the next term of this series will be first we need to check if this particular series is a AP or GP if the series is neither open and terms and check if both the terms of the number individually make a AP or GP and once you know that you can use that to the value of any term at any place in the sequence.

Complete step by step answer:
The sequence given is neither in arithmetic progression or geometric progression that’s why we first divide each term in this sequence where both of the parts individually give us AP and GP with other terms that is

& 2+\dfrac{1}{3},6-\dfrac{1}{6},10+\dfrac{1}{12}..... \\\ & \\\ \end{aligned}$$ So as we can see the first part of terms that is $$2,6,10..$$are in arithmetic progression with the difference between two numbers being $$4$$ . Now to find the sixth term in this sequence we will use the formula of AP which is $${{a}_{n}}={{a}_{1}}+(n-1)d$$ Substituting values we know $${{a}_{6}}=2+(6-1)4$$ Solving the bracket $${{a}_{6}}=2+(5)4$$ Opening the bracket we get the value of the AP part of the sixth term which is $${{a}_{6}}=22$$. Now for the second part of the sequence we can see that the terms are in geometric progression with r being $$r=-\dfrac{1}{2}$$ . Now we can therefore find the sixth term here using the formula we know to find the nth term in a geometric progression that is $${{a}_{n}}={{a}_{1}}{{(r)}^{n-1}}$$ $${{a}_{6}}=\dfrac{1}{3}{{(-\dfrac{1}{2})}^{6-1}}$$ Solving and opening the bracket $${{a}_{6}}=\dfrac{1}{3}\times -\dfrac{1}{32}$$ Therefore the geometric progression part of sixth term is $${{a}_{6}}=-\dfrac{1}{96}$$ Now since we know both part of the second term we add that and get the sixth term therefore $$Sixth\text{ }term=22-\dfrac{1}{96}$$ $$Sixth\text{ }term=\dfrac{2111}{96}$$ Hence now we know the sixth term in the above given question is $$\dfrac{2111}{96}$$ **So, the correct answer is “Option 3”.** **Note:** Arithmetic progression is a sequence of numbers where every successive term is a sum of its preceding term and a fixed difference. Geometric progression is where every term bears a constant ratio to the term before it.