The number of terms common to the two arithme | vedclass

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(D) For the first $A$.$P$.: $a_1 = 3$,$d_1 = 4$. The general term is $T_n = 3 + (n-1)4 = 4n - 1$.For the second $A$.$P$.: $a_2 = 2$,$d_2 = 7$. The gen

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$14$

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(D) For the first $A$.$P$.: $a_1 = 3$,$d_1 = 4$. The general term is $T_n = 3 + (n-1)4 = 4n - 1$.For the second $A$.$P$.: $a_2 = 2$,$d_2 = 7$. The general term is $T_m = 2 + (m-1)7 = 7m - 5$.Equating the terms: $4n - 1 = 7m - 5 \Rightarrow 4n = 7m - 4$.This implies $7m$ must be a multiple of $4$. Since $7$ is not divisible by $4$,$m$ must be a multiple of $4$. Let $m = 4k$.Then $4n = 7(4k) - 4 \Rightarrow n = 7k - 1$.For the first sequence,$T_n \leq 407 \Rightarrow 4n - 1 \leq 407 \Rightarrow 4n \leq 408 \Rightarrow n \leq 102$.Substituting $n = 7k - 1$: $7k - 1 \leq 102 \Rightarrow 7k \leq 103 \Rightarrow k \leq 14.71$.Since $k$ must be a positive integer,the possible values for $k$ are $1, 2, \ldots, 14$.Thus,there are $14$ common terms.

The number of terms common to the two arithmetic progressions $3, 7, 11, \ldots, 407$ and $2, 9, 16, \ldots, 709$ is

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