The number of terms common to the two A.P.'s | vedclass

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(D) The first $A$.$P$. is $A_1: 3, 7, 11, \ldots, 407$. Here,$a_1 = 3$ and $d_1 = 4$. The general term is $T_n = 3 + (n-1)4 = 4n - 1$.The second $A$.$

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(D) The first $A$.$P$. is $A_1: 3, 7, 11, \ldots, 407$. Here,$a_1 = 3$ and $d_1 = 4$. The general term is $T_n = 3 + (n-1)4 = 4n - 1$.The second $A$.$P$. is $A_2: 2, 9, 16, \ldots, 709$. Here,$a_2 = 2$ and $d_2 = 7$. The general term is $T_m = 2 + (m-1)7 = 7m - 5$.For a common term,$4n - 1 = 7m - 5$,which implies $4n = 7m - 4$. This means $7m$ must be a multiple of $4$. Since $7$ is not divisible by $4$,$m$ must be a multiple of $4$. Let $m = 4k$.Substituting $m = 4k$,we get $4n = 7(4k) - 4$,so $n = 7k - 1$.The first common term is for $k=1$,which is $m=4$,$T_4 = 7(4)-5 = 23$.The common difference of the new $A$.$P$. is $	ext{lcm}(4, 7) = 28$.The common terms are $23, 51, 79, \ldots$. The last term must be $\leq 407$.$23 + (N-1)28 \leq 407$$(N-1)28 \leq 384$$N-1 \leq 13.71$$N \leq 14.71$.Thus,the number of common terms is $N = 14$.

The number of terms common to the two $A$.$P$.'s $3, 7, 11, \ldots, 407$ and $2, 9, 16, \ldots, 709$ is

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