Three number series are given below. Find the values of *x*, *y*, and *z* in each series and then determine the correct relationship among them.

Series 1: 102, 107, 117, 134, (x)

Series 2: 130, 115, 135, 110, (y)

Series 3: (z), 80, 120, 300, 1050

Which of the following relations is correct?

Question

Three number series are given below. Find the values of *x*, *y*, and *z* in each series and then determine the correct relationship among them.

Series 1: 102, 107, 117, 134, (x)

Series 2: 130, 115, 135, 110, (y)

Series 3: (z), 80, 120, 300, 1050

Which of the following relations is correct?

Accepted Answer

**Solution:** 1. **Find *x* in Series 1:** 102, 107, 117, 134, (x) - Differences: 107 - 102 = 5 - 117 - 107 = 10 - 134 - 117 = 17 The differences are increasing but not in a simple arithmetic progression. Let's check the pattern: - 5, 10, 17 - Differences between differences: 10 - 5 = 5, 17 - 10 = 7 Since the second difference is increasing by 2, next difference might be 17 + 9 = 26 So, *x* = 134 + 26 = 160 2. **Find *y* in Series 2:** 130, 115, 135, 110, (y) - Differences: 115 - 130 = -15 - 135 - 115 = 20 - 110 - 135 = -25 The differences alternate in sign and increase in magnitude by 5: - -15, +20, -25 Next difference should be +30 So, *y* = 110 + 30 = 140 3. **Find *z* in Series 3:** (z), 80, 120, 300, 1050 - Ratios: - 80 / z - 120 / 80 = 1.5 - 300 / 120 = 2.5 - 1050 / 300 = 3.5 The ratios increase by 1 each time: 1.5, 2.5, 3.5 So, the first ratio (80 / z) should be 0.5 Therefore, $z = \frac{80}{0.5} = 160$ 4. **Compare values:** - $x = 160$ - $y = 140$ - $z = 160$ 5. **Check the options:** - (a) $x > y = z$ → $160 > 140 = 160$ (False, 140 ≠ 160) - (b) $y < x = z$ → $140 < 160 = 160$ (True) - (c) $x = y = z$ → $160 = 140 = 160$ (False) - (d) $y > x > z$ → $140 > 160 > 160$ (False) - (e) $x = z < y$ → $160 = 160 < 140$ (False) **Answer:** (b) $y < x = z$

Answer

Answer

Answer

Answer