The number of terms common between the two se | vedclass

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(B) The first series is $2, 5, 8, \dots$ with $50$ terms.The $n$-th term is $a_n = 2 + (n-1)3 = 3n - 1$. The $50$-th term is $3(50) - 1 = 149$.The sec

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$20$

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(B) The first series is $2, 5, 8, \dots$ with $50$ terms.The $n$-th term is $a_n = 2 + (n-1)3 = 3n - 1$. The $50$-th term is $3(50) - 1 = 149$.The second series is $3, 5, 7, 9, \dots$ with $60$ terms.The $n$-th term is $b_n = 3 + (n-1)2 = 2n + 1$. The $60$-th term is $2(60) + 1 = 121$.Common terms must satisfy $3n - 1 = 2m + 1$,which simplifies to $3n = 2m + 2$,or $3n = 2(m+1)$.This implies $n$ must be even. Let $n = 2k$. Then $3(2k) = 2(m+1) \Rightarrow m = 3k - 1$.The common terms form an arithmetic progression starting at $5$ (where $n=2, m=1$) with a common difference of $6$ ($LCM$ of $3$ and $2$).The common series is $5, 11, 17, \dots, T_k$.We require $T_k \leq 121$ (the limit of the second series).$5 + (k-1)6 \leq 121 \Rightarrow 6k - 1 \leq 121 \Rightarrow 6k \leq 122 \Rightarrow k \leq 20.33$.Since $k$ must be an integer,the maximum value is $k = 20$.

The number of terms common between the two series $2 + 5 + 8 + \dots$ up to $50$ terms and the series $3 + 5 + 7 + 9 + \dots$ up to $60$ terms is:

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