Given the sequence of numbers x1, x2, x3,.......x1005 which satisfy x1x1+1=x2x2+3=x3x3+5=........=x1005x1005+2009 also x1+x2+x3+.....x1005=2010 then 21st term of the sequence is equal to

k=1+(1005)22010 ⇒k=1+10052=10072(k−1−10052) ∴x21+41x21=k⇒x21+41=k x21 ⇒x21=41k−1=41×21005=821005

Given the sequence of numbers x1, x2, x3,.......x1005 which satisfy x1x1+1=x2x2+3=x3x3+5=........=x1005x1005+2009 also x1+x2+x3+.....x1005=2010 then 21st term of the sequence is equal to

k=1+(1005)22010 ⇒k=1+10052=10072(k−1−10052) ∴x21+41x21=k⇒x21+41=k x21 ⇒x21=41k−1=41×21005=821005

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Given the sequence of numbers $x_{1}, x_{2}, x_{3}, ....... x_{1005}$ which satisfy $\frac{x_{1}}{x_{1} + 1} = \frac{x_{2}}{x_{2} + 3} = \frac{x_{3}}{x_{3} + 5} = ........ = \frac{x_{1005}}{x_{1005} + 2009}$ also $x_{1} + x_{2} + x_{3} + ..... x_{1005} = 2010$ then 21^st term of the sequence is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{79}{1005}$

$\frac{82}{1005}$

$\frac{83}{1005}$

$\frac{86}{1005}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$k = 1 + \frac{{(1005)}^{2}}{2010} \Rightarrow k = 1 + \frac{1005}{2} = \frac{1007}{2} (k - 1 - \frac{1005}{2}) ∴ \frac{x_{21} + 41}{x_{21}} = k \Rightarrow x_{21} + 41 = k x_{21} \Rightarrow x_{21} = \frac{41}{k - 1} = \frac{41 \times 2}{1005} = \frac{82}{1005}$