Question

Let a,b,c are the 7th,11th, and 13th terms of non-constant AP. If a,b,c are also in GP, then find $\dfrac{a}{c}$
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: First we’ll write the 7th,11th and 13th terms of the in the general form then, then we’ll substitute the values of a,b,c and a,b,c are also in GP so we’ll use the property of GP to find the relation between the first term of AP and the common difference of AP. After getting the relation we’ll find the ratio $\dfrac{a}{c}$ by substituting the value of unknown terms in the expression to get the answer.

Complete step by step answer:

Given data: $a,b,c$ are ${7^{th}},{11^{th}}\,and\,\,{13^{th}}$ terms of non-constant AP
$a,b,c$ are also in GP
We know that if ${a_1}$ is the first term of an AP and ${d_1}$ is the common difference
Then the term of the AP is given by ${a_1} + (n - 1){d_1}$.
Let the first term of the AP be ‘A’ and the common difference 'd’
$ \Rightarrow {7^{th}}\,term = A + (7 - 1)D$
Substituting the value of ${7^{th}}$ term
$\therefore a = A + 6D$
$ \Rightarrow {11^{th}}\,term = A + (11 - 1)D$
Substituting the value of ${11^{th}}$ term
$\therefore b = A + 10D$
$ \Rightarrow {13^{th}}\,term = A + (13 - 1)D$
Substituting the value of ${13^{th}}$ term
$\therefore c = A + 12D$
We know that if L, M, N are in GP then
$ \Rightarrow {M^2} = LN$
Similarly, $a,b,c$ are also in GP
$ \Rightarrow {b^2} = ac$
Substituting the value of $a,b,c$
$ \Rightarrow {(A + 10d)^2} = (A + 6d)(A + 12d)$
On simplifying the brackets and using ${(x + y)^2} = {x^2} + {y^2} + 2xy$
$ \Rightarrow {A^2} + 100{d^2} + 20Ad = {A^2} + 72{d^2} + 18Ad$
$ \Rightarrow 100{d^2} - 72{d^2} + 20Ad - 18Ad = 0$
Taking $'d'$ common from all the terms
\[ \Rightarrow d\left( {28d + 2A} \right) = 0\]
Either \[d = 0\] or \[28d + 2A = 0\]
Since the AP is non-constant, \[d \ne 0\]
\[\therefore 28d + 2A = 0\]
Simplifying for the value of ‘A’
\[\therefore A = - 14d\]
Now, we have to find $\dfrac{a}{c}$
Substituting the value of ‘a’ and ‘c’
$ \Rightarrow \dfrac{a}{c} = \dfrac{{A + 6d}}{{A + 12d}}$
Substituting the value of ‘A’
$ = \dfrac{{ - 14d + 6d}}{{ - 14d + 12d}}$
$ = \dfrac{{ - 8d}}{{ - 2d}}$
Dividing the numerator and the denominator with ‘-2d’
$ = 4$
Option(D) is correct.

Note: When we get \[d = 0\], we will have to avoid this solution as it is given that the AP is non-constant but some of the students include this in the solution commenting this can also be a case so remember to apply all the data given in the question to get the accurate answer.