Question

find the middle term of the sequence formed by all three digits numbers which leave a reminder 3, when divided by 4. Also find the sum of all numbers on both sides of the middle term separately.

Answer 1

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:

Step-by-step explanation:

Solution :-

The sequence formed by the given numbers is 103, 107, 111, 115, ........., 999.

This is an A.P. in which a = 103, d = 107 - 103 = 4

T(n) = a + (n - 1) d

⇒  103 + (n - 1)4 = 999

⇒  (n - 1)4 = 896

⇒ n - 1 = 224

⇒ n = 224 + 1

⇒ n = 225

Middle term = (n + 1/2)th term = (225 + 1/2)th term = 113th term.

T(113) = (a + 112d) = 103 + 112 × 4 = 551

T(112) = (551 - 4) = 547.

So, we have to find S(112) and [S(225) - S(113)]

Using the formula S(m) = m/2(a + l) for each sum, we get

⇒ S(112) = 112/2(103 + 547)

⇒ S(112) = 112 × 325

⇒ S(112) = 36400

Now, [S(225) - S(113)]

⇒ [S(225) - S(113)] = 225/2(103 + 999) - 113/2(103 + 551)

⇒ [S(225) - S(113)] = (225 × 551) - (113 × 327)

⇒ [S(225) - S(113)] = 123975 - 36951

⇒ [S(225) - S(113)] = 87024.

Hence, Sum of all numbers on LHS of the middle term is 36400 and sum of all numbers on RHS of the middle term is 87024.