what are the first and last terms of an arithmetic sequence when it's arithmetic means are 35, 15 and -15?​

To find the first and last terms of an arithmetic sequence when given its arithmetic means, we can use the properties of an arithmetic progression.Let the arithmetic sequence be $a_1, a_2, a_3, a_4, a_5$.The arithmetic means are the terms between the first and last terms. So, we have:$a_1, {35, 15, -15}, a_5$$a_2 = 35$, $a_3 = 15$, and $a_4 = -15$.Find the common difference ($d$) of the sequence. We can use any two consecutive means:$d = a_3 - a_2$$d = 15 - 35$$d = {-20}$Now that we have the common difference, we can find the first term ($a_1$) and the last term ($a_5$).To find $a_1$:We know that $a_2 = a_1 + d$.So, $35 = a_1 + (-20)$$35 = a_1 - 20$$a_1 = 35 + 20$$a_1 = {55}$To find $a_5$:We know that $a_4 = a_5 - d$.So, $-15 = a_5 - (-20)$$-15 = a_5 + 20$$a_5 = -15 - 20$$a_5 = {-35}$The arithmetic sequence is $55, 35, 15, -15, -35$.Therefore, the first term of the arithmetic sequence is 55 and the last term is -35.[tex] \: [/tex]

Answered by mjPcontiga | 2025-06-27