Question

A cricketer has to score $4500$ run. Let $a_n$ denotes the number of run he scores in the $n^{th}$ match. If $a_1=a_2=......=a_{10}=150$ and $a_{10},a_{11},a_{12}$,... are in A.P. with common difference -2, then find the total number of matches played by him to score 4500 runs

• A. 34
• B. 35
• C. 31
• D. 51

Answer 1

The correct option is A 34

We have $a_1=a_2=...a_9=150,a_{10},a_{11},...a_{n-9}$
Sum$=4500$
$\therefore 9\times 150+\frac{n}{2}[2a\times a_{10}+(n-1)(-2)]=4500$
$\Rightarrow n^2-151n+3150=0$
$\therefore D={\left(151\right)}^2-4\times 3150=102001$
$\Rightarrow n=\frac{151\pm \sqrt{10201}}{2}=126,25$
But $n$ cannot be $126$, as $a_{126}<0$
$\therefore n=25$
$\therefore$ total number$=25+9=34$