find the number of arithmetic means inserted between 8 and 320 given the sum of the second and the sixth mean is 94
Answer (1)
Answer:The number of arithmetic means inserted between 8 and 320, given that the sum of the second and sixth mean is 94, is 31.Step-by-step explanation: * Define the terms: * Let the first term be a_1 = 8. * Let the last term be L = 320. * Let k be the number of arithmetic means inserted. * The total number of terms in the sequence will be n = k + 2. * The arithmetic means are m_1, m_2, \ldots, m_k. * In terms of the arithmetic progression, m_1 = a_2, m_2 = a_3, and so on, up to m_k = a_{k+1}. * The general formula for any term a_j in an arithmetic progression is a_j = a_1 + (j-1)d, where d is the common difference. * Use the given condition about the means: * We are given that the sum of the second mean (m_2) and the sixth mean (m_6) is 94. * This translates to a_3 + a_7 = 94. * Substitute the general formula for a_3 and a_7: (a_1 + (3-1)d) + (a_1 + (7-1)d) = 94 (a_1 + 2d) + (a_1 + 6d) = 94 2a_1 + 8d = 94. * Solve for the common difference (d): * Substitute a_1 = 8 into the equation: 2(8) + 8d = 94 16 + 8d = 94 8d = 94 - 16 8d = 78 d = \frac{78}{8} = \frac{39}{4} = 9.75. * Solve for the number of means (k): * The last term of the sequence is a_n = a_{k+2} = 320. * Using the general formula for the last term: a_{k+2} = a_1 + ((k+2)-1)d 320 = a_1 + (k+1)d. * Substitute a_1 = 8 and d = 9.75: 320 = 8 + (k+1)(9.75) 320 - 8 = (k+1)(9.75) 312 = (k+1)(9.75) k+1 = \frac{312}{9.75} k+1 = \frac{312}{\frac{39}{4}} = 312 \times \frac{4}{39} = 8 \times 4 = 32. * k = 32 - 1 * k = 31.
