JEE Exam > JEE Questions > A sequence of integers a1 + a2 +.... + an sa...
Start Learning for Free
A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......
Correct answer is '4'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an ...
Solution:
Given: an 2 = an 1 − an
Let's find the value of a1000 using the given information.
a1000^2 = a999 - a1000
a999 = a1000^2 + a1000
Similarly, we can find a1001 and a1002.
a1001^2 = a1000 - a1001
a1000 = a1001^2 + a1001
a1002^2 = a1001 - a1002
a1001 = a1002^2 + a1002
Now, we have three equations and three variables. We can solve them to get the values of a1000, a1001, and a1002.
a1000 = -1/2
a1001 = -1/2 + sqrt(3)/2 i
a1002 = -1/2 - sqrt(3)/2 i
Using the given information, we can find the sum of the first 999 terms and the sum of the first 1003 terms.
Sum of the first 999 terms = a1 + a2 + ... + a999 = 1003
Sum of the first 1003 terms = a1 + a2 + ... + a1003 = -999
We can use these two equations to find the sum of the terms from a1001 to a2002.
Sum of the terms from a1001 to a2002 = (a1001 + a1002) + (a1003 + a1004) + ... + (a2001 + a2002)
= (a1001 + a1002) - (a1 + a2 + ... + a1000)
= (sqrt(3)i)/2 - 1003
Now, we can find the sum of the first 2002 terms.
Sum of the first 2002 terms = (a1 + a2 + ... + a1000) + (a1001 + a1002) + (a1003 + a1004) + ... + (a2001 + a2002)
= 1003 - 1/2 + sqrt(3)/2 i + (sqrt(3)i)/2 - 1003 + (-1/2 - sqrt(3)/2 i)
= -2sqrt(3)i
Finally, we can find the sum of digits of the number -2sqrt(3)i.
Sum of digits = Sum of digits of (-2) + Sum of digits of sqrt(3) + Sum of digits of i
= 2 + 3 + 1
= 6
Therefore, the correct answer is '6'.
Given: an 2 = an 1 − an
Let's find the value of a1000 using the given information.
a1000^2 = a999 - a1000
a999 = a1000^2 + a1000
Similarly, we can find a1001 and a1002.
a1001^2 = a1000 - a1001
a1000 = a1001^2 + a1001
a1002^2 = a1001 - a1002
a1001 = a1002^2 + a1002
Now, we have three equations and three variables. We can solve them to get the values of a1000, a1001, and a1002.
a1000 = -1/2
a1001 = -1/2 + sqrt(3)/2 i
a1002 = -1/2 - sqrt(3)/2 i
Using the given information, we can find the sum of the first 999 terms and the sum of the first 1003 terms.
Sum of the first 999 terms = a1 + a2 + ... + a999 = 1003
Sum of the first 1003 terms = a1 + a2 + ... + a1003 = -999
We can use these two equations to find the sum of the terms from a1001 to a2002.
Sum of the terms from a1001 to a2002 = (a1001 + a1002) + (a1003 + a1004) + ... + (a2001 + a2002)
= (a1001 + a1002) - (a1 + a2 + ... + a1000)
= (sqrt(3)i)/2 - 1003
Now, we can find the sum of the first 2002 terms.
Sum of the first 2002 terms = (a1 + a2 + ... + a1000) + (a1001 + a1002) + (a1003 + a1004) + ... + (a2001 + a2002)
= 1003 - 1/2 + sqrt(3)/2 i + (sqrt(3)i)/2 - 1003 + (-1/2 - sqrt(3)/2 i)
= -2sqrt(3)i
Finally, we can find the sum of digits of the number -2sqrt(3)i.
Sum of digits = Sum of digits of (-2) + Sum of digits of sqrt(3) + Sum of digits of i
= 2 + 3 + 1
= 6
Therefore, the correct answer is '6'.
Free Test
FREE
| Start Free Test |
Community Answer
A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an ...
For every integer n ≥ 1
an+3 = an+2 − an+1 = an+1 − an − an+1 = −an
an + 4 = −an+1
an+5 = −an+2
Sum of every six consecutive terms = 0
Let Sn denotes the sum of first 'nn' terms
S999 = S6×166+3 = S3=1003 As every '6' consecutive terms has sum zero.
Hence answer is 4.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer?
Question Description
A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer?.
A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer?.
Solutions for A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer?, a detailed solution for A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? has been provided alongside types of A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A sequence of integers a1 + a2 +.... + an satisfies an+2 = an+1 − an for n ≥ 1. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. If the sum of the first 2002 terms is k, then the sum of digits of number k is......Correct answer is '4'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup