Solveeit Logo

Question

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 \(...

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 ac\dfrac{a}{c}
A. 1
B. 2
C. 3
D. 4

Explanation

Solution

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 ac\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,ca,b,c are 7th,11thand13th{7^{th}},{11^{th}}\,and\,\,{13^{th}} terms of non-constant AP
a,b,ca,b,c are also in GP
We know that if a1{a_1} is the first term of an AP and d1{d_1} is the common difference
Then the term of the AP is given by a1+(n1)d1{a_1} + (n - 1){d_1}.
Let the first term of the AP be ‘A’ and the common difference 'd’
7thterm=A+(71)D\Rightarrow {7^{th}}\,term = A + (7 - 1)D
Substituting the value of 7th{7^{th}} term
a=A+6D\therefore a = A + 6D
11thterm=A+(111)D\Rightarrow {11^{th}}\,term = A + (11 - 1)D
Substituting the value of 11th{11^{th}} term
b=A+10D\therefore b = A + 10D
13thterm=A+(131)D\Rightarrow {13^{th}}\,term = A + (13 - 1)D
Substituting the value of 13th{13^{th}} term
c=A+12D\therefore c = A + 12D
We know that if L, M, N are in GP then
M2=LN\Rightarrow {M^2} = LN
Similarly, a,b,ca,b,c are also in GP
b2=ac\Rightarrow {b^2} = ac
Substituting the value of a,b,ca,b,c
(A+10d)2=(A+6d)(A+12d)\Rightarrow {(A + 10d)^2} = (A + 6d)(A + 12d)
On simplifying the brackets and using (x+y)2=x2+y2+2xy{(x + y)^2} = {x^2} + {y^2} + 2xy
A2+100d2+20Ad=A2+72d2+18Ad\Rightarrow {A^2} + 100{d^2} + 20Ad = {A^2} + 72{d^2} + 18Ad
100d272d2+20Ad18Ad=0\Rightarrow 100{d^2} - 72{d^2} + 20Ad - 18Ad = 0
Taking d'd' common from all the terms
d(28d+2A)=0\Rightarrow d\left( {28d + 2A} \right) = 0
Either d=0d = 0 or 28d+2A=028d + 2A = 0
Since the AP is non-constant, d0d \ne 0
28d+2A=0\therefore 28d + 2A = 0
Simplifying for the value of ‘A’
A=14d\therefore A = - 14d
Now, we have to find ac\dfrac{a}{c}
Substituting the value of ‘a’ and ‘c’
ac=A+6dA+12d\Rightarrow \dfrac{a}{c} = \dfrac{{A + 6d}}{{A + 12d}}
Substituting the value of ‘A’
=14d+6d14d+12d= \dfrac{{ - 14d + 6d}}{{ - 14d + 12d}}
=8d2d= \dfrac{{ - 8d}}{{ - 2d}}
Dividing the numerator and the denominator with ‘-2d’
=4= 4
Option(D) is correct.

Note: When we get d=0d = 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.