The number of common terms in the progression | vedclass

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Question

(C) The first progression is $4, 9, 14, 19, \ldots$ with $a_1 = 4$ and $d_1 = 5$. The $25^{	ext{th}}$ term is $T_{25} = 4 + (25-1)5 = 4 + 120 = 124$.

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) The first progression is $4, 9, 14, 19, \ldots$ with $a_1 = 4$ and $d_1 = 5$. The $25^{	ext{th}}$ term is $T_{25} = 4 + (25-1)5 = 4 + 120 = 124$.The second progression is $3, 6, 9, 12, \ldots$ with $a_2 = 3$ and $d_2 = 3$. The $37^{	ext{th}}$ term is $T_{37} = 3 + (37-1)3 = 3 + 108 = 111$.The common terms must satisfy both progressions. The first common term is $9$.The common difference of the new progression formed by common terms is $	ext{LCM}(d_1, d_2) = 	ext{LCM}(5, 3) = 15$.Let the common terms be $a_n = 9 + (n-1)15$. We need $a_n \le \min(124, 111) = 111$.$9 + (n-1)15 \le 111$$(n-1)15 \le 102$$n-1 \le \frac{102}{15} = 6.8$$n \le 7.8$.Since $n$ must be an integer,the number of common terms is $7$.

The number of common terms in the progressions $4, 9, 14, 19, \ldots$ up to $25^{\text{th}}$ term and $3, 6, 9, 12, \ldots$ up to $37^{\text{th}}$ term is:

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