Q. 104.3( 15 Votes )

How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer.

Answer :

Here, first term = a =63

Common difference = d = 60 - 63 = - 3


Let first n terms of the AP sums to 693.


Sn = 693


To find: n


Now, Sn = (n/2) × [2a + (n - 1)d]


Since, Sn = 693


(n/2) × [2a + (n - 1)d] = 693


(n/2) × [2(63) + (n - 1)(-3)] = 693


(n/2) × [126 - 3n + 3)] = 693


(n/2) × [129 - 3n] = 693


n[129 - 3n] = 1386


129n - 3n2 = 1386


3n2 - 129n + 1386 = 0


(n - 22)( n - 21)= 0


n = 22 or n = 21


n= 22 or n = 21


Since, a22 = a + 21d


= 63 + 21(-3)


= 0


Both the first 21 terms and 22 terms give the sum 693 because the 22nd term is 0. So, the sum doesn’t get affected.


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