Q. 14

Find the middle t

Answer :

The numbers which leave 3 as remainder when divided by 4 will be in the form 3q + 4, for any integer q.


Hence, three digit numbers which leave 3 as remainder when divided by 4 are:


103, 107, 111, …, 199


Clearly above sequence is an AP, with


First term, a = 103


Common difference, d = a2 – a = 107 – 103 = 4


Let the number of terms be ‘n’.


We know, nth term of an AP is


an = a + (n – 1)d


we obtain last term = 199


an = 199


a + (n – 1)d = 199


103 + (n – 1)4 = 199


4(n – 1) = 96


n – 1 = 16


n = 17


Number of terms = 17


Hence, middle term =


and a9 = a + 8d


= 103 + 8(4)


= 135


Hence, middle term of the sequence is 135


Also, we know


We know, sum of ‘n’ terms of an AP is



Now, sum of terms left to the middle term(9th term)Sleft = S8



= 4(206 + 28)


= 936


Now, sum of terms right to the middle (9th term) Sright = S18 – S9



= 2466 – 1071


= 1395


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