Question

The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Answer 1

The general form of an arithmetic sequence is \(a_n = a + (n − 1) d\). Using this formula, the third term \(a_3\) is expressed as \(a_3 = a + (3 − 1) d\), which simplifies to \(a_3 = a + 2d\). Similarly, the seventh term \(a_7\) is expressed as \(a_7 = a + 6d\). It is given that the sum of the third and seventh terms is 6: \(a_3 + a_7 = 6\). Substituting the expressions for \(a_3\) and \(a_7\), we get \((a + 2d) + (a + 6d) = 6\). This simplifies to \(2a + 8d = 6\), and further to \(a + 4d = 3\). From this equation, we can express \(a\) in terms of \(d\): \(a = 3 − 4d\), labeled as equation (i). Additionally, it is given that the product of the third and seventh terms is 8: \((a_3) × (a_7) = 8\). Substituting the expressions for \(a_3\) and \(a_7\), we get \((a + 2d) × (a + 6d) = 8\). Using equation (i), we substitute \(a = 3 − 4d\) into the product equation: \((3-4d+2d)(3-4d+6d) = 8\). This simplifies to \((3-2d)(3+2d) = 8\). Expanding this, we get \(9-4d^2 = 8\). Rearranging the terms, we find \(4d^2 = 9-8\), which means \(4d^2 = 1\). Solving for \(d^2\), we get \(d^2 = \frac 14\). Therefore, the possible values for \(d\) are \(d = ±\frac 12\), meaning \(d = \frac 12\) or \(d = -\frac 12\). Now, we find the corresponding values of \(a\) using equation (i), \(a = 3-4d\). Case 1: When \(d = \frac 12\). \(a = 3-4(\frac 12) = 3-2 = -1\). Case 2: When \(d = -\frac 12\). \(a = 3-4(-\frac 12) = 3+2 = 5\). The formula for the sum of the first \(n\) terms of an arithmetic sequence is \(S_n = \frac n2[2a+(n-1)d]\). We need to calculate \(S_{16}\). Scenario 1: \(a = -1\) and \(d = \frac 12\). (Note: The input text incorrectly stated \(a=1\) here, but based on the calculation that follows, it should be \(a=-1\)) \(S_{16} = \frac {16}{2}[2(-1)+(16-1)(\frac 12)]\) \(S_{16} = 8[-2+\frac {15}{2}]\) \(S_{16} = 8[\frac {-4+15}{2}]\) \(S_{16} = 8 \times \frac {11}{2}\) \(S_{16} = 44\) Scenario 2: \(a = 5\) and \(d = -\frac 12\). \(S_{16} = \frac {16}{2}[2(5)+(16-1)(-\frac 12)]\) \(S_{16} = 8[10+15(-\frac 12)]\) \(S_{16} = 8[10-\frac {15}{2}]\) \(S_{16} = 8[\frac {20-15}{2}]\) \(S_{16} = 8 \times \frac 52\) \(S_{16} = 20\)