Question

Match the following terms such that the arithmetic means of these terms equals the middle term of the AP

• A. $a_{15}$
• B. $a_{10}$
• C. $a_2$

Answer 1

If there are 15 terms in an AP, then the $8^{\text{th}}$ term, $a_8$ will be the middle term of the AP.

This means that $a_8$ will be the arithmetic mean of $a_1$ and $a_{15},$ $a_2$ and $a_{14},$ and so on.

Let us try to solve this in another way.
Let $a$ and $d$ denote the first term and the common difference of the AP respectively.
The $n^{\text{th}}$ term of this AP is given by
$a_n=a+(n-1)d.$
Let $x^{\text{th}}$ term be the required term (such that the aritmetic means of the first and $x^{\text{th}}$term equals the middle term).
$⟹\frac{1}{2}\times (a+x)=a_8$
$⟹\frac{1}{2}\times (a+x)=a+7d$
$⟹x=a+14d=a_{15}$

Similarly,
$\frac{1}{2}\times (a_6+x)=a_8$
$⟹\frac{1}{2}\times (a+5d+x)=a+7d$
$⟹x=a+9d=a_9$

Also,
$\frac{1}{2}\times (a_{14}+x)=a_8$
$⟹\frac{1}{2}\times (a+13d+x)=a+7d$
$⟹x=a+d=a_2$