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Given that Sn = 2n2 − 3n, find S6.

Mathematics

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Answer

B

Step-by-step solution

9 steps

Okay Austino, let's tackle these.

17.

  1. Substitute n=6n=6 into the given formula for SnS_n.

    The formula is Sn=2n2−3nS_n = 2n^2 - 3n.
    We need to find S6S_6.

    S6=2(6)2−3(6)S_6 = 2(6)^2 - 3(6)
  2. Calculate the value.

    S6=2(36)−18S_6 = 2(36) - 18 S6=72−18S_6 = 72 - 18 S6=54S_6 = 54

    The value of S6S_6 is 54.

    The correct option is B.

    B. 54\boxed{\text{B. } 54}

    18.

  3. Define variables and set up the probability equation.

    Let the number of green balls be G=20G = 20.
    Let the number of red balls be RR.
    The total number of balls is T=G+R=20+RT = G + R = 20 + R.
    The probability of selecting a red ball is P(red)=Number of red ballsTotal number of balls=R20+RP(\text{red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{R}{20+R}.
    Given that P(red)=35P(\text{red}) = \frac{3}{5}.
    So, we have the equation:

    R20+R=35\frac{R}{20+R} = \frac{3}{5}
  4. Solve the equation for RR.

    Cross-multiply:

    5R=3(20+R)5R = 3(20+R) 5R=60+3R5R = 60 + 3R

    Subtract 3R3R from both sides:

    5R−3R=605R - 3R = 60 2R=602R = 60

    Divide by 2:

    R=602R = \frac{60}{2} R=30R = 30

    There are 30 red balls.

  5. Calculate the total number of balls in the bag.

    Total balls T=20+R=20+30T = 20 + R = 20 + 30.

    T=50T = 50

    There are 50 balls in the bag.

    The correct option is B.

    B. 50\boxed{\text{B. } 50}

    19.

  6. Write expressions for the terms of the Arithmetic Progression.

    The first term is a=20a = 20.
    The nn-th term of an AP is given by an=a+(n−1)da_n = a + (n-1)d, where dd is the common difference.
    The third term is a3=a+(3−1)d=a+2da_3 = a + (3-1)d = a + 2d.
    The fifth term is a5=a+(5−1)d=a+4da_5 = a + (5-1)d = a + 4d.

  7. Use the given condition to form an equation.

    The fifth term is four-fifth of the third term:

    a5=45a3a_5 = \frac{4}{5} a_3
  8. Substitute the expressions for a3a_3 and a5a_5 into the equation.

    a+4d=45(a+2d)a + 4d = \frac{4}{5}(a + 2d)
  9. Substitute a=20a=20 and solve for dd.

    20+4d=45(20+2d)20 + 4d = \frac{4}{5}(20 + 2d)

    Multiply both sides by 5 to eliminate the fraction:

    5(20+4d)=4(20+2d)5(20 + 4d) = 4(20 + 2d) 100+20d=80+8d100 + 20d = 80 + 8d

    Subtract 8d8d from both sides:

    100+20d−8d=80100 + 20d - 8d = 80 100+12d=80100 + 12d = 80

    Subtract 100 from both sides:

    12d=80−10012d = 80 - 100 12d=−2012d = -20

    Divide by 12:

    d=−2012d = -\frac{20}{12}

    Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 4:

    d=−20÷412÷4d = -\frac{20 \div 4}{12 \div 4} d=−53d = -\frac{5}{3}

    The common difference is ∗−53∗*-\frac{5}{3}*.

    The correct option is C.

    C. −53\boxed{\text{C. } -\frac{5}{3}}
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