How many terms are common in the two arithmetic progressions given bel

Question

How many terms are common in the two arithmetic progressions given below?

AP1: 1, 4, 7, ...,  t_{45} 
AP2: 1, 3, 5, 7, ...,  t_{55}

(A) 15
(B) 16
(C) 18
(D) 19
(E) 22

IanStewart · Answer

It's not necessary, but it's slightly easier to subtract 1 from every term in both sequences - that won't affect how often they overlap. Then we can answer the question using these sequences instead:

0, 3, 6, ... t_45
0, 2, 4, 6, ... t_55

So the first sequence just contains multiples of 3, and the second sequence just contains even numbers (multiples of 2). The numbers that are in both sequences will be multiples of 3 and 2, so will be the multiples of 6 that are small enough to be in both sequences.

The first sequence has 45 terms, so we'll be adding '3' a total of 44 times, and since we're starting from 0, the last term will be (3)(44) = 132.

The second sequence has 55 terms, so we'll be adding '2' a total of 54 times, and since we're starting from 0, the last term will be 108.

So we just need to know how many positive multiples of 6 there are up to 108, and since 108/6 = 18, there are 18 positive multiples of 6 that are in both sequences. There's also the '0' at the start of each sequence that we haven't counted yet, so the answer is 19.

$!vakumar.m · Answer

In the first AP : t1 =1 and d =3.. therefore, t45 = t1+ (45-1)d = 1+44*3 = 133

In the second AP: t1 =1 and d=2..therefore, t55 = 1 + (55-1)*2 = 109

If we note the first few common terms in both the series, it is 1, 7, 13, 19... that is their difference is 6.. hence we need to find the number of such differences/terms upto 109 (being lowest) which are multiples of 6... ie 18 terms (as 18*6 = 108)
and since we found 18 differences, therefore the number of terms is 18+1 = 19 terms
Answer is D

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How many terms are common in the two arithmetic progressions given bel

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How many terms are common in the two arithmetic progressions given bel

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