Question

Each of the series S1=2+4+6+......... and S2=3+6+9+......... is continued to 100 terms. Find how many terms are identical.

• A. 34
• B. 33
• C. 32
• D. None of these

Answer 1

The Correct Option is B

The problem involves finding the number of identical terms in two arithmetic series, S1 and S2, each continued to 100 terms.

First, identify the arithmetic progression (AP) for both series:

• S1: The first term is a₁=2 and the common difference d₁=2. Formula for the n-th term: a_n=a₁+(n−1)d₁=2+(n−1)×2=2n.
• S2: The first term is a₁=3 and the common difference d₂=3. Formula for the n-th term: a_n=a₁+(n−1)d₂=3+(n−1)×3=3n.

The goal is to find how many terms are common in both series.

Since we need S1 terms to equal S2 terms, set 2n=3m, where n and m are term indices in S1 and S2 respectively:

2n=3m ➔ n=(3/2)m ➔ m needs to be even.

S1 n-th term: 2n, with n being an integer such that 1 ≤ n ≤ 100.

If m is even, let m=2k, where k is an integer:

Substitute m in the equation: n=(3/2)(2k)=3k.

We need 3k≤100, hence k≤33.33, meaning the largest integer, k, is 33.

Thus, m takes values: 2, 4, ..., 66 (33 terms).

k	m=2k	n=3k
1	2	3
2	4	6
...	...	...
33	66	99

There are 33 values of both m and n, hence the number of terms identical in both series is 33.