- Home
- Standard 11
- Mathematics
8. Sequences and Series
hard
ધારો કે $x_1, x_2 \ldots, x_{100}$ સમાંતર શ્રેણીમાં છે, જ્યાં $x_1=2$ અને તેઓનો મધ્યક $200$ છે.જો $y_i=i\left(x_i-i\right), 1 \leq i \leq 100$ હોય,તો $y_1, y_2, \ldots, y_{100}$ નો મધ્યક
$..........$ છે.
A
$10101.50$
B
$10051.50$
C
$10049.50$
D
$10100$
(JEE MAIN-2023)
Solution
$\text { Mean }=200$
$\Rightarrow \frac{\frac{100}{2}(2 \times 2+99 d)}{100}=200$
$\Rightarrow 4+99 d =400$
$\Rightarrow d=4$
$y_i=i(x i-i)$
$=i(2+(i-1) 4-i)=3 i^2-2 i$
$\text { Mean }=\frac{\sum y_i}{100}$
$=\frac{1}{100} \sum \limits_{i=1}^{100} 3 i^2-2 i$
$=\frac{1}{100}\left\{\frac{3 \times 100 \times 101 \times 201}{6}-\frac{2 \times 100 \times 101}{2}\right\}$
$=101\left\{\frac{201}{2}-1\right\}=101 \times 99.5$
$=10049 \cdot 50$
Standard 11
Mathematics
