Let a, b and c be the 7^{th}, 11^{th} and 13^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the first term of the $A.P.$ be $A$ and the common difference be $d$. Since the $A.P.$ is non-constant,$d 
eq 0$.Given terms are:$a = A + 6d$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let the first term of the $A.P.$ be $A$ and the common difference be $d$. Since the $A.P.$ is non-constant,$d 
eq 0$.Given terms are:$a = A + 6d$$b = A + 10d$$c = A + 12d$Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.$(A + 10d)^2 = (A + 6d)(A + 12d)$$A^2 + 20Ad + 100d^2 = A^2 + 18Ad + 72d^2$$2Ad = -28d^2$Since $d 
eq 0$,we divide by $2d$ to get $A = -14d$,or $\frac{A}{d} = -14$.Now,we calculate $\frac{a}{c}$:$\frac{a}{c} = \frac{A + 6d}{A + 12d}$Dividing numerator and denominator by $d$:$\frac{a}{c} = \frac{\frac{A}{d} + 6}{\frac{A}{d} + 12} = \frac{-14 + 6}{-14 + 12} = \frac{-8}{-2} = 4$.

Let $a, b$ and $c$ be the $7^{th}, 11^{th}$ and $13^{th}$ terms respectively of a non-constant $A.P.$ If these are also the three consecutive terms of a $G.P.$,then $\frac{a}{c}$ is equal to

Similar Questions

If ${a_1}, {a_2}, {a_3}, \dots, {a_n}$ are in $H.P.$,then ${a_1}{a_2} + {a_2}{a_3} + \dots + {a_{n-1}}{a_n}$ is equal to:

If $x, y, z$ are in $H.P.$,then the value of the expression $\log(x + z) + \log(x - 2y + z)$ is

Find the sum of the first hundred even natural numbers divisible by $5$.

If the ${5^{th}}$ term of a $H.P.$ is $\frac{1}{45}$ and the ${11^{th}}$ term is $\frac{1}{69}$,then its ${16^{th}}$ term will be:

The value of $\sum\limits_{n = 2}^\infty {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module