Question

What is the number common term in the two arithmetic progression: 13,15,17 and 14,17,20 and if each progression has 100 terms?

Answer 1

Answer:

33 terms.

Step-by-step explanation:

By inference it is clear that 1st AP has common difference 2 and second one has 3.

r₁th term be any term in first AP and r₂th term be that of second.

So r₁ th term is 13+(r₁-1)2  and r₂ th term is 14 + (r₂-1)3

They must be equal

i.e 13 + (r₁-1)2 = 14 + (r₂-1)3

2r₁ -2 = 1 + 3r₂ - 3

2r₁=3r₂

r₁/r₂ = 3/2⇒ r₁ = 3k and r₂ = 2k where k is a natural number.

But there are only 100 terms in each AP i.e 1 < r₁,r₂<100,

Consider  r₁ = 3k

1<3k<100

1/3 < k < 100/3 ..........(1)

Similarly for r₂ = 2k

1/2 < K < 50......(2)

From (1) and (2),

1/2 < k < 100/3

0.5 < k < 33.33.

Since k is a natural number,

it takes values 1,2,3,4,5,...,33

So 33 terms are common in those APs.