Answer to Question #230716 in Calculus for Paa

Question

Let {a_n} be an arithmetic sequence such that its 1^st, 20^th, and 58^th terms are consecutive terms of some geometric sequence. Find the common ratio of the geometric sequence.

Answer 1

Solution:

$a_1,a_{20},a_{58}$ are in GP.

Let the first term $a_1$ be $a$ and the common difference is $d$.

$a_{20}=a+19d \\a_{58}=a+57d$

When p,q,r are in GP, then $q^2=pr$

So, $a_{20}^2=a_1\times a_{58}$

$\Rightarrow (a+19d)^2=a(a+57d) \\ \Rightarrow a^2+361d^2+38ad=a^2+57ad \\ \Rightarrow 361d^2-19ad=0 \\\Rightarrow 19d(19d-a)=0 \\\Rightarrow d=0;d=\dfrac{a}{19}$

Rejecting d=0 as it will give trivial AP.

So, $a_1=a,a_{20}=a+19(\dfrac{a}{19})=2a,a_{58}=a+57(\dfrac{a}{19})=4a$

Now, a, 2a, 4a are in GP.

Common ratio$=r=\dfrac{2a}{a}=2$