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Given a_n = 5+3(n−1).

Mathematics

ScanSolve AI Answer

Answer

3
75

Step-by-step solution

6 steps
  1. Identify the common difference.

    The nthn^{th} term of an arithmetic progression is given by an=a+(n−1)da_n = a + (n-1)d, where aa is the first term and dd is the common difference.
    Given an=5+3(n−1)a_n = 5+3(n-1).
    Comparing this to the general formula, we can see that the common difference dd is the coefficient of (n−1)(n-1).

    d=3d = 3
  2. Find the first term (a1a_1) and the sixth term (a6a_6).

    Substitute n=1n=1 into the given formula to find the first term:

    a1=5+3(1−1)=5+3(0)=5a_1 = 5+3(1-1) = 5+3(0) = 5

    Substitute n=6n=6 into the given formula to find the sixth term:

    a6=5+3(6−1)=5+3(5)=5+15=20a_6 = 5+3(6-1) = 5+3(5) = 5+15 = 20
  3. Calculate the sum of the first six terms (S6S_6).

    The formula for the sum of the first nn terms of an A.P. is Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n).
    For n=6n=6:

    S6=62(a1+a6)S_6 = \frac{6}{2}(a_1 + a_6) S6=3(5+20)S_6 = 3(5 + 20) S6=3(25)S_6 = 3(25) S6=75S_6 = 75

    The common difference is 3 and the sum of the first six terms is 75.

    The common difference is 3\boxed{3}
    The sum of the first six terms is 75\boxed{75}

  4. Rationalize the first term.

    Multiply the numerator and denominator by the conjugate of the denominator, 5−35-\sqrt{3}.

    35+3=35+3×5−35−3\frac{\sqrt{3}}{5+\sqrt{3}} = \frac{\sqrt{3}}{5+\sqrt{3}} \times \frac{5-\sqrt{3}}{5-\sqrt{3}} =3(5−3)(5)2−(3)2= \frac{\sqrt{3}(5-\sqrt{3})}{(5)^2 - (\sqrt{3})^2} =53−(3)225−3= \frac{5\sqrt{3} - (\sqrt{3})^2}{25 - 3} =53−322= \frac{5\sqrt{3} - 3}{22}
  5. Rationalize the second term.

    Multiply the numerator and denominator by the conjugate of the denominator, 5+3\sqrt{5}+\sqrt{3}.

    255−3=255−3×5+35+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}} =25(5+3)(5)2−(3)2= \frac{2\sqrt{5}(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2} =2(5)2+2535−3= \frac{2(\sqrt{5})^2 + 2\sqrt{5}\sqrt{3}}{5 - 3} =2(5)+2152= \frac{2(5) + 2\sqrt{15}}{2} =10+2152= \frac{10 + 2\sqrt{15}}{2} =5+15= 5 + \sqrt{15}
  6. Add the rationalized terms.

    53−322+(5+15)\frac{5\sqrt{3} - 3}{22} + (5 + \sqrt{15})

    To combine, find a common denominator:

    =53−322+22(5+15)22= \frac{5\sqrt{3} - 3}{22} + \frac{22(5 + \sqrt{15})}{22} =53−3+110+221522= \frac{5\sqrt{3} - 3 + 110 + 22\sqrt{15}}{22} =107+53+221522= \frac{107 + 5\sqrt{3} + 22\sqrt{15}}{22}

    This expression is not in the form a+bca+b\sqrt{c} with a single c\sqrt{c} term. Let's recheck the problem statement. It asks for a+bca+b\sqrt{c}. This implies cc should be a single integer. The current result has 3\sqrt{3} and 15\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)+215=10+215-2(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)
    15−3+515=13−3+10+215\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+315-7+3\sqrt{15}
    2

    Let's re-evaluate my calculation for the second term:
    255−3=25(5+3)(5−3)(5+3)=2(5)+2155−3=10+2152=5+15\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:
    35+3=3(5−3)(5+3)(5−3)=53−325−3=53−322\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 53−322+5+15\frac{5\sqrt{3}-3}{22} + 5+\sqrt{15}.
    This is −322+5322+5+15\frac{-3}{22} + \frac{5\sqrt{3}}{22} + 5 + \sqrt{15}.
    This cannot be simplified into the form a+bca+b\sqrt{c} where cc is a single integer, unless 3\sqrt{3} and 15\sqrt{15} can be combined, which they cannot.
    Perhaps the question implies that cc 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 35−3\frac{\sqrt{3}}{5-\sqrt{3}} or the second term was 235−3\frac{2\sqrt{3}}{\sqrt{5}-\sqrt{3}}, it might simplify differently.
    Given the problem as written: 35+3+255−3=a+bc\frac{\sqrt{3}}{5+\sqrt{3}} + \frac{2\sqrt{5}}{\sqrt{5}-\sqrt{3}} = a+b\sqrt{c}.

    Let's assume cc is meant to be 15 or 3.
    If c=15c=15, then we need to combine terms with 15\sqrt{15}.
    If c=3c=3, then we need to combine terms with 3\sqrt{3}.

    Let's look at the student's work again. The student has a term −7+3152\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.

    53−322+22(5+15)22=53−3+110+221522=107+53+221522\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} =10722+5223+222215= \frac{107}{22} + \frac{5}{22}\sqrt{3} + \frac{22}{22}\sqrt{15} =10722+5223+15= \frac{107}{22} + \frac{5}{22}\sqrt{3} + \sqrt{15}

    This is of the form a+b1c1+b2c2a+b_1\sqrt{c_1}+b_2\sqrt{c_2}, not a+bca+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 cc is a specific value, it's not stated.
    If cc can be either 3 or 15, then the expression is not uniquely a+bca+b\sqrt{c}.

    Let's consider the possibility that the question expects cc 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: 35+3=3(5−3)25−3=53−322\frac{\sqrt{3}}{5+\sqrt{3}} = \frac{\sqrt{3}(5-\sqrt{3})}{25-3} = \frac{5\sqrt{3}-3}{22}
    Term 2: 255−3=25(5+3)5−3=10+2152=5+15\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: 53−322+5+15\frac{5\sqrt{3}-3}{22} + 5+\sqrt{15}
    =−322+5322+5+15= \frac{-3}{22} + \frac{5\sqrt{3}}{22} + 5 + \sqrt{15}
    =(5−322)+5322+15= \left(5 - \frac{3}{22}\right) + \frac{5\sqrt{3}}{22} + \sqrt{15}
    =(110−322)+5322+15= \left(\frac{110-3}{22}\right) + \frac{5\sqrt{3}}{22} + \sqrt{15}
    =10722+5223+15= \frac{107}{22} + \frac{5}{22}\sqrt{3} + \sqrt{15}

    This result has two different square root terms, 3\sqrt{3} and 15\sqrt{15}.
    If the question implies that cc is a single integer, then the problem as stated cannot be simplified to a+bca+b\sqrt{c}.
    However, if cc is not specified, then the expression is in the form a+b1c1+b2c2a+b_1\sqrt{c_1}+b_2\sqrt{c_2}.
    Given the format a+bca+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 3\sqrt{3} or 5\sqrt{5} should be different to allow for simplification.
    For example, if the first term was 55+3\frac{\sqrt{5}}{5+\sqrt{3}}, it would still lead to mixed radicals.
    If the second term was 235−3\frac{2\sqrt{3}}{\sqrt{5}-\sqrt{3}}, then:
    23(5+3)5−3=215+2(3)2=15+3\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 53−322+15+3=53−3+2215+6622=63+53+221522\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 cc is not an integer? The problem states a,b,ca, b, c are constants. Usually, cc 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 −7+3152\frac{-7+3\sqrt{15}}{2}.
    If we assume the final form is a+b15a+b\sqrt{15}, then the 3\sqrt{3} term must cancel or be zero. This is not the case.

    Given the exact wording "find the value of aa and bb", and the form a+bca+b\sqrt{c}, it is highly likely that cc 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 35+3+255−3=a+bc\frac{\sqrt{3}}{5+\sqrt{3}} + \frac{2\sqrt{5}}{\sqrt{5}-\sqrt{3}} = a+b\sqrt{c}, where aa, bb and cc are constants, find the value of aa and bb."
    It does not ask for cc. It asks for aa and bb. This implies cc is implicitly determined by the expression.
    If the expression results in 10722+5223+15\frac{107}{22} + \frac{5}{22}\sqrt{3} + \sqrt{15}, then it cannot be written as a+bca+b\sqrt{c} for a single cc.

    Could it be that cc is a product of primes, e.g., c=3×5=15c=3 \times 5 = 15?
    If c=15c=15, then we need to express 3\sqrt{3} in terms of 15\sqrt{15}. 3=155\sqrt{3} = \frac{\sqrt{15}}{\sqrt{5}}. This doesn't help.

    Let's assume the

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