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

The correct option is B $42$

A.M.

$a,b,c,d,e,f:AM$ to be inserted in b/w 2& 12

$\therefore$ Since, in a series : $A,a,b,c,d,e,f,B$

$\Rightarrow A+B=a+f=b+e=c+a$

$\therefore$ In the given series : $2,a,b,c,d,e,f,12$

$2+12=(a+f)=(b+e)=(c+d)=14...\left(1\right)$

$\therefore$ To find $:a+b+c+d+e+f$

$\Rightarrow (a+f)+(b+e)+(d+c)$

$\Rightarrow 14+14+14$ (from (1))

$\Rightarrow 14\times 3$

$\Rightarrow 42$ (option B)