To solve 2x + 3 > 7, subtract 3 from both sides: 2x > 4. Then divide by 2: x > 2.
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Q2. Which expression represents the sum of the first n terms of an arithmetic series with first term a and common difference d?
Solution:
The formula for the sum of the first n terms is S_n = n/2 * (2a + (n-1)d).
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Q3. Solve the inequality 3x - 5 < 7.
Solution:
To solve 3x - 5 < 7, add 5 to both sides: 3x < 12. Then divide by 3: x < 4.
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Q4. Factor the polynomial x^2 - 9.
Solution:
The expression x^2 - 9 is a difference of squares, which factors to (x - 3)(x + 3).
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Q5. If the first term of a geometric progression is 5 and the common ratio is 3, what is the 3rd term?
Solution:
The nth term of a geometric progression is given by a_n = a * r^(n-1). Here, a = 5, r = 3, and n = 3. So, a_3 = 5 * 3^(3-1) = 5 * 3^2 = 5 * 9 = 45.
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Q6. If the first term of a geometric progression is 3 and the common ratio is 2, what is the 4th term?
Solution:
The nth term of a geometric progression is given by a_n = a * r^(n-1). Here, a = 3, r = 2, and n = 4. So, a_4 = 3 * 2^(4-1) = 3 * 2^3 = 3 * 8 = 24.
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Q7. What is the sum of the first 5 terms of the arithmetic progression 2, 5, 8, ...?
Solution:
The first term a = 2, common difference d = 3. The sum of the first n terms S_n = n/2 * (2a + (n-1)d). For n = 5, S_5 = 5/2 * (2*2 + 4*3) = 5/2 * (4 + 12) = 5/2 * 16 = 40/2 = 20.
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Q8. What is the sum of the first 6 terms of the arithmetic progression 1, 4, 7, ...?
Solution:
The first term a = 1, common difference d = 3. The sum of the first n terms S_n = n/2 * (2a + (n-1)d). For n = 6, S_6 = 6/2 * (2*1 + 5*3) = 3 * (2 + 15) = 3 * 17 = 51.
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Q9. Solve for x: 5x + 3 = 2x + 12.
Solution:
First, subtract 2x from both sides: 3x + 3 = 12. Then subtract 3: 3x = 9. Finally, divide by 3: x = 3.
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Q10. What is the value of x in the equation x^2 + 6x + 9 = 0?
Solution:
This is a perfect square trinomial: (x + 3)^2 = 0. Therefore, x + 3 = 0, which gives x = -3.
