Given a sequence of 4 numbers,the first three | vedclass

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

Question

(B) Let the sequence be $a, b, c, d$.Since the first three terms $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since the last three terms $b, c, d$ are i

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let the sequence be $a, b, c, d$.Since the first three terms $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since the last three terms $b, c, d$ are in $A.P.$ with a common difference of $6$,we have $c - b = 6$ and $d - c = 6$.Given that the first and last terms are equal,$a = d$.From $d - c = 6$,we have $d = c + 6$. Since $a = d$,then $a = c + 6$,which implies $c = a - 6$.From $c - b = 6$,we have $b = c - 6 = (a - 6) - 6 = a - 12$.Substituting $b = a - 12$ and $c = a - 6$ into the $G.P.$ condition $b^2 = ac$:$(a - 12)^2 = a(a - 6)$$a^2 - 24a + 144 = a^2 - 6a$$144 = 18a$$a = 8$.Since $a = d$,the last term $d = 8$.

Given a sequence of $4$ numbers,the first three of which are in $G.P.$ and the last three are in $A.P.$ with a common difference of $6$. If the first and last terms in this sequence are equal,then the last term is:

Similar Questions

How many terms of the $A.P.$ $1, 4, 7, \ldots$ are needed to give the sum $715$?

Let $a_1, a_2, ..., a_{10}$ be a $G.P.$ If $\frac{a_3}{a_1} = 25$,then $\frac{a_9}{a_5}$ is equal to:

$\prod\limits_{n = 1}^{10} {\left( {\frac{{\left( {6\sum\limits_{i = 0}^n i } \right) + 1}}{{\left( {6\sum\limits_{j = 0}^n {(j - 1)} } \right) + 1}}} \right)} $ is equal to:

For $x \in R$,let $[x]$ denote the greatest integer $\le x$. Find the sum of the series $\left[ -\frac{1}{3} \right] + \left[ -\frac{1}{3} - \frac{1}{100} \right] + \left[ -\frac{1}{3} - \frac{2}{100} \right] + \dots + \left[ -\frac{1}{3} - \frac{99}{100} \right]$.

If the $(p + q)^{th}$ term of a $G.P.$ is $m$ and the $(p - q)^{th}$ term is $n$,then the $p^{th}$ term will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module