Question

In the expansion of $(x+y{)}^{25}$,
$1^{st}$ term from the end = $(26-p{)}^{th}$ term from the beginning
$2^{nd}$ term from the end = $(26-q{)}^{th}$ term from the beginning
$3^{rd}$ term from the end = $(26-r{)}^{th}$ term from the beginning
${10}^{st}$ term from the end = $(26-s{)}^{th}$ term from the beginning
Find the value of p+q+r+s

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Answer 1

There will be 26 terms in the expansion of $(x+y{)}^{25}$
Therefore the $1^{st}$ term from the end = $(26{)}^{th}$ term from the beginning
$\Rightarrow p=0$
Similarly, $2^{nd}$ term from the end =$(25{)}^{th}$ term from the beginning
$\Rightarrow q=1$
When we expend this, we get to see that $3^{rd}$ term from the end = $(26-2{)}^{th}$ term from the beginning and ${10}^{th}$ term from the end = (26-9)th term from the beginning.
$\Rightarrow p=0,q=1,r=2ands=9$
$\Rightarrow p+q+r+s=12$