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Find the 21st term of the A.P. – 4, – 1.5, 1, 3.5,...

Mathematics

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Answer

B

46

Step-by-step solution

55 steps

Here are the solutions to questions 14 to 30:

Part 14· 3 steps

Find the 21st21^{st} term of the A.P -4, -1.5, 1, 3.5 ...

  1. Identify the first term (aa) and the common difference (dd).

    $$
a = -4
$$
$$
d = -1.5 - (-4) = -1.5 + 4 = 2.5
$$

  2. Use the formula for the nthn^{th} term of an A.P., Tn=a+(n−1)dT_n = a + (n-1)d.

    We need T21T_{21}. T21=−4+(21−1)(2.5) T_{21} = -4 + (21-1)(2.5) T21=−4+(20)(2.5) T_{21} = -4 + (20)(2.5)

  3. Calculate the value.

    $$
T_{21} = -4 + 50
$$
$$
T_{21} = 46
$$
    • The correct answer is (b) 46.

Part 15· 3 steps

Simplify 32−22\sqrt{32} - 2\sqrt{2}

  1. Simplify 32\sqrt{32}.

    $$
\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}
$$

  2. Substitute the simplified form back into the expression.

    $$
4\sqrt{2} - 2\sqrt{2}
$$

  3. Combine the terms.

    $$
(4-2)\sqrt{2} = 2\sqrt{2}
$$
    • The correct answer is (a) 222\sqrt{2}.

Part 16· 4 steps

The 3rd3^{rd} and 7th7^{th} term of a G.P. are 81 and 16. Find the 5th5^{th} term.

  1. For a Geometric Progression (GP), the nthn^{th} term is Tn=arn−1T_n = ar^{n-1}.

    $$
T_3 = ar^2 = 81 \quad (\text{Equation 1})
$$
$$
T_7 = ar^6 = 16 \quad (\text{Equation 2})
$$

  2. Divide Equation 2 by Equation 1 to find the common ratio rr.

    $$
\frac{ar^6}{ar^2} = \frac{16}{81}
$$
$$
r^4 = \frac{16}{81}
$$
$$
r = \sqrt[4]{\frac{16}{81}} = \frac{2}{3}
$$

  3. Substitute r=23r = \frac{2}{3} into Equation 1 to find the first term aa.

    $$
a\left(\frac{2}{3}\right)^2 = 81
$$
$$
a\left(\frac{4}{9}\right) = 81
$$
$$
a = \frac{81 \times 9}{4} = \frac{729}{4}
$$

  4. Calculate the 5th5^{th} term, T5=ar4T_5 = ar^4.

    $$
T_5 = \frac{729}{4} \times \left(\frac{2}{3}\right)^4
$$
$$
T_5 = \frac{729}{4} \times \frac{16}{81}
$$
$$
T_5 = \frac{729}{81} \times \frac{16}{4}
$$
$$
T_5 = 9 \times 4 = 36
$$
    • The correct answer is (d) 36.

Part 17· 3 steps

If sin⁡θ=3/5\sin \theta = ^3/_5, 0∘<θ<90∘0^\circ < \theta < 90^\circ, Find cos⁡θ\cos \theta.

  1. Use the trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.

    $$
\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1
$$

  2. Solve for cos⁡2θ\cos^2 \theta.

    $$
\frac{9}{25} + \cos^2 \theta = 1
$$
$$
\cos^2 \theta = 1 - \frac{9}{25}
$$
$$
\cos^2 \theta = \frac{25 - 9}{25} = \frac{16}{25}
$$

  3. Take the square root.

    Since 0∘<θ<90∘0^\circ < \theta < 90^\circ, cos⁡θ\cos \theta is positive. cos⁡θ=1625=45 \cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}

    • The correct answer is (a) 4/5^4/_5.

Part 18· 2 steps

Factorize x2−4x−21x^2 - 4x - 21

  1. Find two numbers that multiply to -21 and add up to -4.

    The numbers are -7 and 3 (since $-7 \times 3 = -21$ and $-7 + 3 = -4$).

  2. Write the quadratic expression in factored form.

    $$
(x-7)(x+3)
$$
    • The correct answer is (d) (x−7)(x+3)(x-7)(x+3).

Part 19· 3 steps

Solve the quadratic equation 2x2−3x−5=02x^2 - 3x - 5 = 0

  1. Factorize the quadratic equation.

    Find two numbers that multiply to 2×(−5)=−102 \times (-5) = -10 and add to -3. These numbers are -5 and 2. 2x2−5x+2x−5=0 2x^2 - 5x + 2x - 5 = 0

  2. Group terms and factor out common factors.

    $$
x(2x - 5) + 1(2x - 5) = 0
$$
$$
(x+1)(2x-5) = 0
$$

  3. Set each factor to zero and solve for xx.

    $$
x+1 = 0 \implies x = -1
$$
$$
2x-5 = 0 \implies 2x = 5 \implies x = \frac{5}{2}
$$
    • The correct answer is (a) −1-1 or 5/2^5/_2.

Part 20· 2 steps

Express 0.000724 in standard form

  1. Move the decimal point to the right until there is one non-zero digit before it.

    $$
0.000724 \implies 7.24
$$

  2. Count the number of places the decimal point moved.

    It moved 4 places to the right. Since it moved to the right, the exponent will be negative. 7.24×10−4 7.24 \times 10^{-4}

    • The correct answer is (c) 7.24×10−47.24 \times 10^{-4}.

Part 21· 4 steps

Solve the inequality −2(13+x)≥9+5x-2(13+x) \ge 9+5x.

  1. Distribute the -2 on the left side.

    $$
-26 - 2x \ge 9 + 5x
$$

  2. Gather xx terms on one side and constant terms on the other.

    Add 2x2x to both sides. −26≥9+5x+2x -26 \ge 9 + 5x + 2x −26≥9+7x -26 \ge 9 + 7x

  3. Subtract 9 from both sides.

    $$
-26 - 9 \ge 7x
$$
$$
-35 \ge 7x
$$

  4. Divide by 7.

    The inequality sign does not change because we are dividing by a positive number. −357≥x \frac{-35}{7} \ge x −5≥x -5 \ge x

    • This can also be written as x≤−5x \le -5.
    • The correct answer is (d) x≤−5x \le -5.

Part 22· 3 steps

Express 98\sqrt{98} in basic form

  1. Find the largest perfect square factor of 98.

    $$
98 = 49 \times 2
$$

  2. Rewrite the square root using this factor.

    $$
\sqrt{98} = \sqrt{49 \times 2} = \sqrt{49} \times \sqrt{2}
$$

  3. Simplify the perfect square.

    $$
7\sqrt{2}
$$
    • The correct answer is (b) 727\sqrt{2}.

Part 23· 3 steps

Express 288\sqrt{288} in basic form

  1. Find the largest perfect square factor of 288.

    $$
288 = 144 \times 2
$$

  2. Rewrite the square root using this factor.

    $$
\sqrt{288} = \sqrt{144 \times 2} = \sqrt{144} \times \sqrt{2}
$$

  3. Simplify the perfect square.

    $$
12\sqrt{2}
$$
    • The correct answer is (b) 12212\sqrt{2}.

Part 24· 5 steps

Simplify 12+4824\frac{\sqrt{12} + \sqrt{48}}{\sqrt{24}}

  1. Simplify each surd in the expression.

    $$
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}
$$
$$
\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}
$$
$$
\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}
$$

  2. Substitute the simplified surds back into the expression.

    $$
\frac{2\sqrt{3} + 4\sqrt{3}}{2\sqrt{6}} = \frac{6\sqrt{3}}{2\sqrt{6}}
$$

  3. Simplify the fraction.

    $$
\frac{6\sqrt{3}}{2\sqrt{6}} = \frac{3\sqrt{3}}{\sqrt{6}}
$$

  4. Rationalize the denominator by multiplying the numerator and denominator by 6\sqrt{6}.

    $$
\frac{3\sqrt{3}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{18}}{6}
$$

  5. Simplify 18\sqrt{18} and then the entire expression.

    $$
\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}
$$
$$
\frac{3 \times 3\sqrt{2}}{6} = \frac{9\sqrt{2}}{6} = \frac{3\sqrt{2}}{2}
$$
    • The correct answer is (d) 322\frac{3\sqrt{2}}{2}.

Part 25· 2 steps

Simplify 9a5×4a39a^5 \times 4a^3

  1. Multiply the coefficients and combine the variables using the exponent rule am×an=am+na^m \times a^n = a^{m+n}.

    $$
(9 \times 4) \times (a^5 \times a^3)
$$
$$
36 \times a^{5+3}
$$

  2. Calculate the final expression.

    $$
36a^8
$$
    • The correct answer is (c) 36a836a^8.

Part 26· 3 steps

Simplify log⁡412−log⁡436+log⁡4192\log_4 12 - \log_4 36 + \log_4 192

  1. Use the logarithm properties log⁡bx−log⁡by=log⁡b(xy)\log_b x - \log_b y = \log_b (\frac{x}{y}) and log⁡bx+log⁡by=log⁡b(xy)\log_b x + \log_b y = \log_b (xy).

    $$
\log_4 12 - \log_4 36 + \log_4 192 = \log_4 \left(\frac{12}{36}\right) + \log_4 192
$$

  2. Simplify the fraction and then combine the remaining terms.

    $$
= \log_4 \left(\frac{1}{3}\right) + \log_4 192
$$
$$
= \log_4 \left(\frac{1}{3} \times 192\right)
$$
$$
= \log_4 \left(\frac{192}{3}\right)
$$
$$
= \log_4 64
$$

  3. Determine the value of log⁡464\log_4 64.

    We need to find xx such that 4x=644^x = 64. 4x=43 4^x = 4^3 x=3 x = 3

    • The correct answer is (a) 3.

Part 27· 3 steps

Simplify 3n+1×31−n3^{n+1} \times 3^{1-n}

  1. Use the exponent rule am×an=am+na^m \times a^n = a^{m+n}.

    $$
3^{(n+1) + (1-n)}
$$

  2. Simplify the exponent.

    $$
3^{n+1+1-n} = 3^2
$$

  3. Calculate the value.

    $$
3^2 = 9
$$
    • The correct answer is (d) 9.

Part 28· 2 steps

Solve 4x=644^x = 64

  1. Express both sides of the equation with the same base.

    $$
4^x = 4^3
$$

  2. Equate the exponents.

    $$
x = 3
$$
    • The correct answer is (c) 3.

Part 29· 5 steps

Solve 9x+1=81x−29^{x+1} = 81^{x-2}

  1. Express both sides of the equation with the same base.

    Since 81=9281 = 9^2, we can use base 9. 9x+1=(92)x−2 9^{x+1} = (9^2)^{x-2}

  2. Apply the exponent rule (am)n=amn(a^m)^n = a^{mn}.

    $$
9^{x+1} = 9^{2(x-2)}
$$
$$
9^{x+1} = 9^{2x-4}
$$

  3. Equate the exponents.

    $$
x+1 = 2x-4
$$

  4. Solve for xx.

    Subtract xx from both sides. 1=x−4 1 = x-4

  5. Add 4 to both sides.

    $$
1+4 = x
$$
$$
x = 5
$$
    • The correct answer is (b) 5.

Part 30· 5 steps

Given that 183y=2y+2\frac{1}{8^{3y}} = 2^{y+2}, find y.

  1. Express both sides of the equation with the same base.

    The smallest common base is 2. 8=23 8 = 2^3 1(23)3y=2y+2 \frac{1}{(2^3)^{3y}} = 2^{y+2}

  2. Apply the exponent rules (am)n=amn(a^m)^n = a^{mn} and 1am=a−m\frac{1}{a^m} = a^{-m}.

    $$
\frac{1}{2^{9y}} = 2^{y+2}
$$
$$
2^{-9y} = 2^{y+2}
$$

  3. Equate the exponents since the bases are the same.

    $$
-9y = y+2
$$

  4. Solve for yy.

    Subtract yy from both sides. −9y−y=2 -9y - y = 2 −10y=2 -10y = 2

  5. Divide by -10.

    $$
y = \frac{2}{-10}
$$
$$
y = -\frac{1}{5}
$$
    • My calculation yields y=−15y = -\frac{1}{5}. This answer is not among the provided options (a) 1, (b) 2, (c) 3, (d) 4. There might be a typo in the question or the options. However, if we assume the question intended to be 8y=2y+28^y = 2^{y+2} (a common type of simplification error in such problems), then: 23y=2y+22^{3y} = 2^{y+2} 3y=y+23y = y+2 2y=22y = 2 y=1y = 1
    • Assuming this interpretation, the answer would be (a) 1.
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