Question
The number of persons in the groups are chosen as $7,11,18,29,47,76,123,199 \ldots .$ the number of persons in the $(\mathrm{n}-2)$ th, nth and $(\mathrm{n}+1)$ th groups $(\mathrm{n} \geq 3)$, form(a) a GP(b) a HP(c) an AP(d) an AGP
Step 1
$ We can see that each term is the sum of the previous two terms. This means that the sequence is a Fibonacci sequence. Show more…
Show all steps
Your feedback will help us improve your experience
Gaurav Kalra and 87 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
There are $n$ persons $(n \geq 3)$, among whom are $A$ and $B$, who are made to stand in a row in random order. Probability that there is exactly one person between $A$ and $B$ is (A) $\frac{n-2}{n(n-1)}$ (B) $\frac{2(n-2)}{n(n-1)}$ (C) $2 / n$ (D) none os these
If $a_{1}, a_{2}, a_{3} \ldots$ are in HP and $f(k)=\sum_{r=1}^{n} a_{r}-a_{k}$, then $\frac{a_{1}}{f(1)}, \frac{a_{2}}{f(2)}, \frac{a_{3}}{f(3)}, \ldots \frac{a=}{f(n)}$ are in (a) AP (b) GP (c) HP (d) AGP
If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in $\mathrm{AP}$ and $\mathrm{a}, \mathrm{mb}, \mathrm{c}$ are in GP, then $\mathrm{a}, \mathrm{m}^{2} \mathrm{~b}, \mathrm{c}$ are in (a) AP (b) $\mathrm{GP}$ (c) HP (d) AGP
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD