Find the arithmetic sequence whose fifth term \( =21 \) and its tenth term equals to three times of its second term

To derive the arithmetic sequence, let's denote the first term as \( a \) and the common difference as \( d \). From the information given: 1. The fifth term can be expressed as \( a + 4d = 21 \). 2. The tenth term can be expressed as \( a + 9d \) and is equal to three times the second term \( a + d \). Therefore, we have: \[ a + 9d = 3(a + d) \] Simplifying this equation gives: \[ a + 9d = 3a + 3d \] \[ 6d = 2a \] \[ a = 3d \] Now we can substitute \( a \) back into the first equation: \[ 3d + 4d = 21 \] \[ 7d = 21 \] \[ d = 3 \] Now substituting \( d \) back into the equation for \( a \): \[ a = 3 \cdot 3 = 9 \] Thus, the first term is \( 9 \) and the common difference is \( 3 \). The arithmetic sequence is given by: \[ 9, 12, 15, 18, 21, 24, 27, \ldots \] This sequence follows the formula \( a_n = a + (n-1)d \) where \( a = 9 \) and \( d = 3 \).