Question

If sum of 101 distinct terms in AP is zero, in how many ways can three of these terms be selected such that the sum is zero

Answer 1

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There are two Arithmetic Progressions A and B such that their nth terms are given by An = 101 + 3(n 1) and Bn = 150 + (n 1), where n is the set of natural numbers. The first 50 terms of A and B are written alternately i.e. A1B1A2B2..A50B50. What is the remainder when the number so formed is divided by 11?

a1 b 0 c 9 d 10

cl mock test question i guess ..

A1 = 101 A50 = 248

B1 = 150 B50 = 199

now in groups of 3 we add -- B50 --- B1

and A50 ---A1

{25(150 + 199) - 25(101 + 248 )} (mod 11)

0 is the answer