Question

If $a$, $b$, $c$, $d$, $e$, $f$ are A.M’s between $2$ and $12$, then $a+b+c+d+e+f$ is equal to
A. $14$
B. $42$
C. $84$
D. None of these

Answer 1

In this problem we need to calculate the value of $a+b+c+d+e+f$, where $a$, $b$, $c$, $d$, $e$, $f$ are A.M’s between $2$ and $12$. We know that if $a$, $b$, $c$, $d$, $e$, $f$ are A.M’s between $2$ and $12$, then the variables $a$, $b$, $c$, $d$, $e$, $f$ forms a arithmetic progression(A.P) with $2$ as first term and $12$ as last term. Now we will write the sum of the $n$ terms of A.P with first term ${{a}_{0}}$ and last term ${{a}_{n}}$ as ${{S}_{n}}=\dfrac{n}{2}\left( {{a}_{0}}+{{a}_{n}} \right)$. From this formula we will calculate the sum of the all terms in the obtained A.P. after that we will sum all the terms in the A.P and equate it to the calculated sum and simplify It by using mathematical operations to get the required result.

Complete step by step answer:
Given that $a$, $b$, $c$, $d$, $e$, $f$ are A.M’s between $2$ and $12$. So the terms $a$, $b$, $c$, $d$, $e$, $f$ forms an Arithmetic progression A.P with $2$ as first term and $12$ as last term. Which is given by
$2$, $a$, $b$, $c$, $d$, $e$, $f$, $12$.
In the above A.P we have $8$ terms, so $n=8$.
Now the sum of the $n$ terms in A.P with first term ${{a}_{0}}$ and last term ${{a}_{n}}$ as ${{S}_{n}}=\dfrac{n}{2}\left( {{a}_{0}}+{{a}_{n}} \right)$. From this formula the sum of $8$ terms of the obtained A.P is given by
${{S}_{8}}=\dfrac{8}{2}\left( 2+12 \right)$
But the sum of the $8$ will be $2+a+b+c+d+e+f+12$. Substituting this value in the above equation, then we will get
$2+a+b+c+d+e+f+12=\dfrac{8}{2}\left( 2+12 \right)$
Simplifying the above equation by using basic mathematical operations, then we will have
$\begin{aligned} & 14+a+b+c+d+e+f=4\times 14 \\\ & \Rightarrow 14+a+b+c+d+e+f=56 \\\ & \Rightarrow a+b+c+d+e+f=56-14 \\\ & \Rightarrow a+b+c+d+e+f=42 \\\ \end{aligned}$

So, the correct answer is “Option B”.

Note: We can also solve this problem in another manner. If $a$, $b$, $c$, $d$, $e$, $f$ are A.M’s between $2$ and $12$, then $2+12=a+f=b+e=c+d$. From this equation we can calculate the values of $a+f$, $b+e$, $c+d$. Now add all the values and simplify it to get the required result.