If the arithmetic mean between the p^{th} ter | vedclass

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(A) Let the arithmetic progression be $a, a+d, a+2d, \dots$ where $a$ is the first term and $d$ is the common difference.The $n^{th}$ term is given by

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$r + s$

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(A) Let the arithmetic progression be $a, a+d, a+2d, \dots$ where $a$ is the first term and $d$ is the common difference.The $n^{th}$ term is given by $T_n = a + (n-1)d$.The arithmetic mean between $p^{th}$ and $q^{th}$ terms is $\frac{T_p + T_q}{2} = \frac{(a + (p-1)d) + (a + (q-1)d)}{2} = \frac{2a + (p+q-2)d}{2} = a + \frac{(p+q-2)d}{2}$.The arithmetic mean between $r^{th}$ and $s^{th}$ terms is $\frac{T_r + T_s}{2} = \frac{(a + (r-1)d) + (a + (s-1)d)}{2} = \frac{2a + (r+s-2)d}{2} = a + \frac{(r+s-2)d}{2}$.Given that these means are equal:$a + \frac{(p+q-2)d}{2} = a + \frac{(r+s-2)d}{2}$.Subtracting $a$ from both sides and dividing by $\frac{d}{2}$ (assuming $d 
eq 0$):$p + q - 2 = r + s - 2$.Therefore,$p + q = r + s$.

If the arithmetic mean between the $p^{th}$ term and $q^{th}$ term of an arithmetic progression is equal to the arithmetic mean between its $r^{th}$ and $s^{th}$ term,then $p + q = ......$

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