Question

The Sum of three numbers in AP is $75$, and product of extremities is $609$. The AM of first and second number is

• A. $\{21,25,29\}$, AM $=23$
• B. $\{13,17,21\}$, AM $=22$
• C. $\{21,25,29\}$, AM $=25$
• D. $\{21,22,29\}$, AM $=23$

Answer 1

The correct option is A $\{21,25,29\}$, AM $=23$

Let the numbers in A.P. be (a-d), a, (a+d)

$\therefore (a-d)+a+(a+d)=75$

$\Rightarrow a=25$

Also, $(a-d)(a+d)=609$

$\Rightarrow a^2-d^2=609$

$\Rightarrow {25}^2-d^2=609$

$\therefore d=\pm 4$

The numbers are {21,25,29} or {29,25,21}

According to option, we take {21,25,29}

A.M. of first and second number $=\frac{21+25}{2}=23$