Welcome back Debby — missed you this week. Step 1: Identify the common difference. The $n^{th}$ term of an arithmetic progression is given by $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference. Given $a_n = 5+3(n-1)$. Comparing this to the general formula, we can see that the common difference $d$ is the coefficient of $(n-1)$. $$d = 3$$ Step 2: Find the first term ($a_1$) and the sixth term ($a_6$). Substitute $n=1$ into the given formula to find the first term: $$a_1 = 5+3(1-1) = 5+3(0) = 5$$ Substitute $n=6$ into the given formula to find the sixth term: $$a_6 = 5+3(6-1) = 5+3(5) = 5+15 = 20$$ Step 3: Calculate the sum of the first six terms ($S_6$). The formula for the sum of the first $n$ terms of an A.P. is $S_n = \frac{n}{2}(a_1 + a_n)$. For $n=6$: $$S_6 = \frac{6}{2}(a_1 + a_6)$$ $$S_6 = 3(5 + 20)$$ $$S_6 = 3(25)$$ $$S_6 = 75$$ The common difference is $3$ and the sum of the first six terms is $75$. The common difference is $\boxed{3}$ The sum of the first six terms is $\boxed{75}$ Step 1: Rationalize the first term. Multiply the numerator and denominator by the conjugate of the denominator, $5-\sqrt{3}$. $$\frac{\sqrt{3}}{5+\sqrt{3}} = \frac{\sqrt{3}}{5+\sqrt{3}} \times \frac{5-\sqrt{3}}{5-\sqrt{3}}$$ $$= \frac{\sqrt{3}(5-\sqrt{3})}{(5)^2 - (\sqrt{3})^2}$$ $$= \frac{5\sqrt{3} - (\sqrt{3})^2}{25 - 3}$$ $$= \frac{5\sqrt{3} - 3}{22}$$ Step 2: Rationalize the second term. Multiply the numerator and denominator by the conjugate of the denominator, $\sqrt{5}+\sqrt{3}$. $$\frac{2\sqrt{5}}{\sqrt{5}-\sqrt{3}} = \frac{2\sqrt{5}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$$ $$= \frac{2\sqrt{5}(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2}$$ $$= \frac{2(\sqrt{5})^2 + 2\sqrt{5}\sqrt{3}}{5 - 3}$$ $$= \frac{2(5) + 2\sqrt{15}}{2}$$ $$= \frac{10 + 2\sqrt{15}}{2}$$ $$= 5 + \sqrt{15}$$ Step 3: Add the rationalized terms. $$\frac{5\sqrt{3} - 3}{22} + (5 + \sqrt{15})$$ To combine, find a common denominator: $$= \frac{5\sqrt{3} - 3}{22} + \frac{22(5 + \sqrt{15})}{22}$$ $$= \frac{5\sqrt{3} - 3 + 110 + 22\sqrt{15}}{22}$$ $$= \frac{107 + 5\sqrt{3} + 22\sqrt{15}}{22}$$ This expression is not in the form $a+b\sqrt{c}$ with a single $\sqrt{c}$ term. Let's recheck the problem statement. It asks for $a+b\sqrt{c}$. This implies $c$ should be a single integer. The current result has $\sqrt{3}$ and $\sqrt{15}$. This suggests there might be a simplification or an error in my calculation or interpretation. Let's re-examine the student's work on the right side of the image. The student has: $-2(5) + 2\sqrt{15} = 10 + 2\sqrt{15}$ (This seems to be from the second term's numerator, but the sign is wrong for the first part) $\sqrt{15-3} + 5\sqrt{15} = \sqrt{13} - 3 + 10 + 2\sqrt{15}$ (This looks like an attempt to combine terms, but it's very messy) $-7+3\sqrt{15}$ $2$ Let's re-evaluate my calculation for the second term: $\frac{2\sqrt{5}}{\sqrt{5}-\sqrt{3}} = \frac{2\sqrt{5}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{2(5) + 2\sqrt{15}}{5-3} = \frac{10+2\sqrt{15}}{2} = 5+\sqrt{15}$. This is correct. Let's re-evaluate my calculation for the first term: $\frac{\sqrt{3}}{5+\sqrt{3}} = \frac{\sqrt{3}(5-\sqrt{3})}{(5+\sqrt{3})(5-\sqrt{3})} = \frac{5\sqrt{3}-3}{25-3} = \frac{5\sqrt{3}-3}{22}$. This is also correct. So, the sum is $\frac{5\sqrt{3}-3}{22} + 5+\sqrt{15}$. This is $\frac{-3}{22} + \frac{5\sqrt{3}}{22} + 5 + \sqrt{15}$. This cannot be simplified into the form $a+b\sqrt{c}$ where $c$ is a single integer, unless $\sqrt{3}$ and $\sqrt{15}$ can be combined, which they cannot. Perhaps the question implies that $c$ is a specific value, or that one of the radical terms should disappear. Let's check if there's a typo in the question. If the first term was $\frac{\sqrt{3}}{5-\sqrt{3}}$ or the second term was $\frac{2\sqrt{3}}{\sqrt{5}-\sqrt{3}}$, it might simplify differently. Given the problem as written: $\frac{\sqrt{3}}{5+\sqrt{3}} + \frac{2\sqrt{5}}{\sqrt{5}-\sqrt{3}} = a+b\sqrt{c}$. Let's assume $c$ is meant to be $15$ or $3$. If $c=15$, then we need to combine terms with $\sqrt{15}$. If $c=3$, then we need to combine terms with $\sqrt{3}$. Let's look at the student's work again. The student has a term $\frac{-7+3\sqrt{15}}{2}$. This suggests they might have gotten a common denominator of 2 for some part. Let's try to get a common denominator of 22 for the entire expression. $$ \frac{5\sqrt{3}-3}{22} + \frac{22(5+\sqrt{15})}{22} = \frac{5\sqrt{3}-3+110+22\sqrt{15}}{22} = \frac{107+5\sqrt{3}+22\sqrt{15}}{22} $$ $$ = \frac{107}{22} + \frac{5}{22}\sqrt{3} + \frac{22}{22}\sqrt{15} $$ $$ = \frac{107}{22} + \frac{5}{22}\sqrt{3} + \sqrt{15} $$ This is of the form $a+b_1\sqrt{c_1}+b_2\sqrt{c_2}$, not $a+b\sqrt{c}$. There might be an error in the problem statement or a misunderstanding of the expected form. However, I must solve the question as given. If the question implies that $c$ is a specific value, it's not stated. If $c$ can be either $3$ or $15$, then the expression is not uniquely $a+b\sqrt{c}$. Let's consider the possibility that the question expects $c$ to be the same for both terms after some manipulation, or that one of the radical terms cancels out. This is not happening here. Let's re-check the arithmetic carefully. Term 1: $\frac{\sqrt{3}}{5+\sqrt{3}} = \frac{\sqrt{3}(5-\sqrt{3})}{25-3} = \frac{5\sqrt{3}-3}{22}$ Term 2: $\frac{2\sqrt{5}}{\sqrt{5}-\sqrt{3}} = \frac{2\sqrt{5}(\sqrt{5}+\sqrt{3})}{5-3} = \frac{10+2\sqrt{15}}{2} = 5+\sqrt{15}$ Sum: $\frac{5\sqrt{3}-3}{22} + 5+\sqrt{15}$ $= \frac{-3}{22} + \frac{5\sqrt{3}}{22} + 5 + \sqrt{15}$ $= \left(5 - \frac{3}{22}\right) + \frac{5\sqrt{3}}{22} + \sqrt{15}$ $= \left(\frac{110-3}{22}\right) + \frac{5\sqrt{3}}{22} + \sqrt{15}$ $= \frac{107}{22} + \frac{5}{22}\sqrt{3} + \sqrt{15}$ This result has two different square root terms, $\sqrt{3}$ and $\sqrt{15}$. If the question implies that $c$ is a single integer, then the problem as stated cannot be simplified to $a+b\sqrt{c}$. However, if $c$ is not specified, then the expression is in the form $a+b_1\sqrt{c_1}+b_2\sqrt{c_2}$. Given the format $a+b\sqrt{c}$, it usually implies a single radical term. Let's assume there might be a typo in the question and one of the $\sqrt{3}$ or $\sqrt{5}$ should be different to allow for simplification. For example, if the first term was $\frac{\sqrt{5}}{5+\sqrt{3}}$, it would still lead to mixed radicals. If the second term was $\frac{2\sqrt{3}}{\sqrt{5}-\sqrt{3}}$, then: $\frac{2\sqrt{3}(\sqrt{5}+\sqrt{3})}{5-3} = \frac{2\sqrt{15}+2(3)}{2} = \sqrt{15}+3$. Then the sum would be $\frac{5\sqrt{3}-3}{22} + \sqrt{15}+3 = \frac{5\sqrt{3}-3+22\sqrt{15}+66}{22} = \frac{63+5\sqrt{3}+22\sqrt{15}}{22}$. Still two radicals. What if $c$ is not an integer? The problem states $a, b, c$ are constants. Usually, $c$ is an integer in these types of problems. Let's consider the possibility that the student's work on the right side is a hint. The student has $\frac{-7+3\sqrt{15}}{2}$. If we assume the final form is $a+b\sqrt{15}$, then the $\sqrt{3}$ term must cancel or be zero. This is not the case. Given the exact wording "find the value of $a$ and $b$", and the form $a+b\sqrt{c}$, it is highly likely that $c$ is a specific integer, and the expression should simplify to have only one radical. Since it doesn't, I will state the result as is, and mention the issue. However, the instructions say "Solve questions completely and accurately." and "Formula → substitute → simplify step by step. Show every step." My steps are accurate for the given expression. Let's re-read the question carefully: "Given that $\frac{\sqrt{3}}{5+\sqrt{3}} + \frac{2\sqrt{5}}{\sqrt{5}-\sqrt{3}} = a+b\sqrt{c}$, where $a$, $b$ and $c$ are constants, find the value of $a$ and $b$." It does not ask for $c$. It asks for $a$ and $b$. This implies $c$ is implicitly determined by the expression. If the expression results in $\frac{107}{22} + \frac{5}{22}\sqrt{3} + \sqrt{15}$, then it cannot be written as $a+b\sqrt{c}$ for a single $c$. Could it be that $c$ is a product of primes, e.g., $c=3 \times 5 = 15$? If $c=15$, then we need to express $\sqrt{3}$ in terms of $\sqrt{15}$. $\sqrt{3} = \frac{\sqrt{15}}{\sqrt{5}}$. This doesn't help. Let's assume the