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

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Question

Let $a,b,c,d$ be four number of the sequence.

Now, according to the question ${b^2} = ac$ and $c-b=6$ and $a-c=6$

Also, given $\boxed{a = d

Vedclass · Accepted Answer

$8$

Vedclass · Answer

Let $a,b,c,d$ be four number of the sequence.

Now, according to the question ${b^2} = ac$ and $c-b=6$ and $a-c=6$

Also, given $\boxed{a = d}$

$	herefore {b^2} = ac \Rightarrow {b^2} = a\left[ {\frac{{a + b}}{2}} ight]$

                                                     ($\because $     $2c=a+b$)

$ \Rightarrow {a^2} - 2{b^2} + ab = 0$

Now, $c-b=6$ and $a-c=6$

givea $a-b=12$

$ \Rightarrow b = a - 12$

$	herefore {a^2} - 2{b^2} + ab = 0$

$ \Rightarrow {a^2} - 2{\left( {a - 12} ight)^2} + a\left( {a - 12} ight) = 0$

$ \Rightarrow {a^2} - 2{a^2} - 288 + 48a + {a^2} - 12a = 0$

$ \Rightarrow 36a = 288 \Rightarrow a = 8$

Hence, last term is $d=a=8$.

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

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