The increasing sequence 3, 15, 24, 48, consists of those positive multiples of 3 that are one less than a perfect square. What is the remainder when 2021st term of the sequence is divided by 1000?
Answers
SOLUTION
GIVEN
The increasing sequence 3, 15, 24, 48,... consists of those positive multiples of 3 that are one less than a perfect square.
TO DETERMINE
The remainder when 2021 st term of the sequence is divided by 1000
EVALUATION
Here the given increasing sequence is
3, 15, 24, 48, ...
So general term of the sequence is n² - 1
Now 3 divides n² - 1
⇒ 3 divides (n +1) (n - 1)
Since 3 is a prime number
So Either 3 divides (n +1) or 3 divides (n - 1)
When 3 divides (n +1) then
n = 3k - 1 where k = 1 , 2 , 3 ,..
When 3 divides (n - 1) then
n = 3k + 1 where k = 1 , 2 , 3 ,..
For n = 3k - 1 we get
n² - 1
= ( 3k - 1) ² - 1
= 9k² - 6k
Which gives the the terms at the odd places of the increasing sequence 3, 15, 24, 48,...
For n = 3k + 1 we get
n² - 1
= ( 3k + 1) ² - 1
= 9k² + 6k
Which gives the the terms at the even places of the increasing sequence 3, 15, 24, 48,...
We first find 2021 st term of the sequence
Since 2021 is odd
So 2021 is a term of the sequence 3k - 1 where k = 1 , 2 , 3 ,..
Solving we get 2021 is 1011 th term of the sequence 3k - 1 where k = 1 , 2 , 3 ,..
Consequently
n² - 1
= ( 3 × 1011 - 1)² - 1
= ( 3033 - 1)² - 1
= 3032² - 1
Now 3032 ≡ 32 ( mod 1000 )
⇒ 3032² ≡ 32² ( mod 1000 )
⇒ 3032² ≡ 1024 ( mod 1000 )
⇒ 3032² - 1 ≡ 1023 ( mod 1000 )
Now 1023 ≡ 23 ( mod 1000 )
Thus we get 3032² - 1 ≡ 23 ( mod 1000 )
Therefore n² - 1 ≡ 23 ( mod 1000 )
FINAL ANSWER
Hence the required Remainder = 23
━━━━━━━━━━━━━━━━
Learn more from Brainly :-
If the middle term of a finite AP with 7 terms is 21 find the sum of all terms of the AP
https://brainly.in/question/30198388
2. find the 100th term of an AP whose nth term is 3n+1
https://brainly.in/question/22293445
3. 2,7,12,17,...sum of 12 terms of this A. P. is
https://brainly.in/question/24134044
Answer:
Step-by-step explanation: