C. Find S10 if5. a₁ = 103and a10 = 36.a₁28and a10 =917.a3 =57 and a7 = 69
Answer (1)
5. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & a_1 = 103 \\ & a_{10} = 3 \\ & n = 10 \end{aligned}[/tex]Step 2: Solve for the sum of the terms of an arithmetic sequence.[tex]\begin{aligned} S_n & = \frac{n}{2}(a_1 + a_n) \\ S_{10} & = \frac{10}{2}(103 + a_{10}) \\ & = 5(103 + 3) \\ & = 5(106) \\ & = \boxed{530} \end{aligned}[/tex]6. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & a_1 = 28 \\ & a_{10} = 91 \\ & n = 10 \end{aligned}[/tex]Step 2: Solve for the sum of the terms of an arithmetic sequence.[tex]\begin{aligned} S_n & = \frac{n}{2}(a_1 + a_n) \\ S_{10} & = \frac{10}{2}(28 + a_{10}) \\ & = 5(28 + 91) \\ & = 5(119) \\ & = \boxed{595} \end{aligned}[/tex]7. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & n = 10 \\ & a_3 = 57 \\ & a_7 = 69 \end{aligned}[/tex]Step 2: Find the common difference.In this case, we will use 7 and 3 as the values of n and m, respectively.[tex]\begin{aligned} a_n & = a_m + (n - m)d \\ a_7 & = a_3 + (7 - 3)d \\ 69 & = 57 + 4d \\ 57 + 4d & = 69 \\ 4d & = 69 - 57 \\ 4d & = 12 \\ \frac{4d}{4} & = \frac{12}{4} \\ d & = 3 \end{aligned}[/tex]Step 3: Get the first term.In this case, we will use 3 as the value of n.[tex]\begin{aligned} a_n & = a_1 + (n - 1)d \\ a_3 & = a_1 + (3 - 1)(3) \\ 57 & = a_1 + (2)(3) \\ 57 & = a_1 + 6 \\ a_1 + 6 & = 57 \\ a_1 & = 57 - 6 \\ a_1 & = 51 \end{aligned}[/tex]Step 4: Solve for the sum of the terms of an arithmetic sequence.We will now use 10 as the value of n.[tex]\begin{aligned} S_n & = \frac{n}{2}[2a_1 + (n - 1)d] \\ S_{10} & = \frac{10}{2}[2(51) + (10 - 1)(3)] \\ & = 5[(102 + (9)(3)] \\ & = 5(102 + 27) \\ & = 5(129) \\ & = \boxed{645} \end{aligned}[/tex]
