Find the middle term of the sequence formed by all three digit numbers which leave a remainder 3. When divided by 4. Also, find the sum of all numbers on both sides of the middle term separately.

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87024

82704

84207

87402

answer is A.

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Detailed Solution

Complete step-by-step answer:
Find the three digit numbers which leaves a remainder 3 when divided by 4
Try diving numbers starting from 100 to 999 and get the required numbers.
The resultant numbers are –
103,107,111,....999
Now, we will find the total number of numbers in the above series.
Here,
a=103
d=107−103=4
an=999
Place the above values in the equation - an=a+(n−1)d
⇒999=103+(n−1)4
Simplify the above equation –
⇒999−103=(n−1)4
⇒896=(n−1)4
It can be re-written as –
⇒(n−1)4=896
⇒(n−1)=
⇒(n−1)=224
Make “n” the subject. When any term is moved from one side to another, then the sign also changes. Positive terms become negative and vice-versa.
⇒n=224+1
⇒n=225
Since the total numbers are odd, middle term can be calculated as –
M=
⇒M=
⇒M=
⇒M=113
Now, find the middle term using an=a+(n−1)d
Where n=113
⇒an=103+(113−1)4
Simplify the above equation –
⇒an=103+(112)4
⇒an=103+448
⇒an=551
Now, sum of all the terms before middle term is
Sn =[2a+(n−1)d]
Here n=112 place the values
⇒Sn=[2(103)+(112−1)4]
Simplify the above equation –
⇒Sn=[206+(111)4]
⇒Sn=56[206+444]⇒Sn=56[650]
⇒Sn=36,400 .... (A)
Similarly, sum of all the terms is
Sn =[2a+(n−1)d]
Here n=225 place the values
⇒Sn=[2(103)+(225−1)4]
Simplify the above equation –
⇒Sn=[206+(224)4]
⇒Sn= [206+896]
⇒Sn= [1102]
⇒Sn=225[551]
⇒Sn=123975 .... (B)
Sum of all the terms after middle terms can be given as –
From the Summation of all the terms subtract the terms before middle term and the middle term
=123975−36400−551
Simplify the above equation –
=87024
So, the correct answer is
“87024”.

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