Question

write the sequence of all three digit numbers which when divided by 4leave reminder2.what is common difference​

Answer 1

Answer

The list of 3 digit number that leaves a remainder of 3 when divided by 4 is :

103  ,  107 , 111 , 115 ,   .... 999  

The above list is in AP with first term, a = 103 and common difference, d = 4

Let n be the number of terms in the AP.

Now, an = 999

103 + ( n  -  1 ) 4 =  999

103 + 4n  - 4 =  999

4n  + 99 = 999

4n  =  900

n  =  225  

Since, the number of terms is odd, so there will be only one middle term.

middle term = (n+12)th term = 113th term = a + 112d = 103 + 112×4 = 551  

We know that, sum of first n terms of an AP is,Sn = n2[2a+(n−1)d]  

Now, Sum = 112/2[2×103 + 111×4] = 36400  

Sum of all terms before  middle term  = 36400

sum of all numbers=  225/2[2×103+224×4] = 123975

Now, sum of terms after  middle term = S225 − (S112+551) = 123975−(36400+551) = 87024

Answer 2

Answer:

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Step-by-step explanation:

The list of 3 digit number that leaves a remainder of 3 when divided by 4 is :

103 , 107 , 111 , 115 , .... 999

The above list is in AP with first term, a = 103 and common difference, d = 4

Let n be the number of terms in the AP.

Now, an = 999

103 + ( n - 1 ) 4 = 999

103 + 4n - 4 = 999

4n + 99 = 999

4n = 900

n = 225

Since, the number of terms is odd, so there will be only one middle term.

middle term = (n+12)th term = 113th term = a + 112d = 103 + 112×4 = 551

We know that, sum of first n terms of an AP is,Sn = n2[2a+(n−1)d]

Now, Sum = 112/2[2×103 + 111×4] = 36400

Sum of all terms before middle term = 36400

sum of all numbers= 225/2[2×103+224×4] = 123975

Now, sum of terms after middle term = S225 − (S112+551) = 123975−(36400+551) = 87024