Two series I and II are given below, each containing one wrong term. Both series follow different patterns. Identify the difference between the wrong term of series II and the correct term of series I at the position of the wrong term.

Series I: $a$, $b+5$, $c+8$, $87$, $412$, $2185$, $13326$

Series II: $b+4$, $c+8$, $36$, $a+50$, $79$, $111$, $152$

Given:

1. The quadratic equation $10b x^2 - (21 - a)x + 3 = 0$ has roots $\frac{z}{10}$ and $\frac{y}{20}$, where $\frac{z}{10} \frac{y}{20}$.
2. Any multiple of $y$ ends with digit zero.
3. $c - z = 9$ and $ ext{LCM}(z, 5) = 15$.

Find the difference between the wrong term of series II and the correct term of series I at the position of the wrong term.

Question

Two series I and II are given below, each containing one wrong term. Both series follow different patterns. Identify the difference between the wrong term of series II and the correct term of series I at the position of the wrong term.

Series I: $a$, $b+5$, $c+8$, $87$, $412$, $2185$, $13326$

Series II: $b+4$, $c+8$, $36$, $a+50$, $79$, $111$, $152$

Given:

1. The quadratic equation $10b x^2 - (21 - a)x + 3 = 0$ has roots $\frac{z}{10}$ and $\frac{y}{20}$, where $\frac{z}{10} > \frac{y}{20}$.
2. Any multiple of $y$ ends with digit zero.
3. $c - z = 9$ and $	ext{LCM}(z, 5) = 15$.

Find the difference between the wrong term of series II and the correct term of series I at the position of the wrong term.

Accepted Answer

**Solution:**

1. From the quadratic equation:
   $$10b x^2 - (21 - a)x + 3 = 0$$
   Roots are given as $\frac{z}{10}$ and $\frac{y}{20}$, with $\frac{z}{10} > \frac{y}{20}$.

2. Using Vieta's formulas:
   - Sum of roots:
     $$\frac{z}{10} + \frac{y}{20} = \frac{21 - a}{10b}$$
   - Product of roots:
     $$\frac{z}{10} 	imes \frac{y}{20} = \frac{3}{10b}$$

3. From the product:
   $$\frac{zy}{200} = \frac{3}{10b} \Rightarrow zy = \frac{600}{b}$$

4. Given $c - z = 9$ and LCM$(z, 5) = 15$.
   Since LCM$(z, 5) = 15$, and 5 is prime, $z$ must be 3 (because LCM(3,5) = 15).

5. Therefore, $c = z + 9 = 3 + 9 = 12$.

6. Since any multiple of $y$ ends with zero, $y$ must be 10 or a factor of 10.

7. Using product relation:
   $$zy = \frac{600}{b} \Rightarrow 3y = \frac{600}{b} \Rightarrow y = \frac{600}{3b} = \frac{200}{b}$$

8. Since $y$ is a factor of 10, possible values for $y$ are 10 or 5 or 2 or 1.
   Checking $y=10$:
   $$10 = \frac{200}{b} \Rightarrow b = 20$$

9. Using sum of roots:
   $$\frac{z}{10} + \frac{y}{20} = \frac{21 - a}{10b}$$
   Substitute $z=3$, $y=10$, $b=20$:
   $$\frac{3}{10} + \frac{10}{20} = \frac{21 - a}{200}$$
   $$0.3 + 0.5 = \frac{21 - a}{200}$$
   $$0.8 = \frac{21 - a}{200}$$
   $$21 - a = 160$$
   $$a = 21 - 160 = -139$$

10. Now, the terms:
    - $a = -139$
    - $b = 20$
    - $c = 12$

11. Check series I terms:
    - $a = -139$
    - $b + 5 = 20 + 5 = 25$
    - $c + 8 = 12 + 8 = 20$
    - Next terms: 87, 412, 2185, 13326

12. Check series II terms:
    - $b + 4 = 20 + 4 = 24$
    - $c + 8 = 12 + 8 = 20$
    - $36$
    - $a + 50 = -139 + 50 = -89$
    - $79, 111, 152$

13. Identify wrong terms:
    - In series I, the term $87$ seems out of pattern (since previous terms are small, next jumps to 87, then 412, etc.).
    - In series II, the term $36$ is suspicious because the pattern around it does not fit.

14. The wrong term in series II is $36$.

15. The correct term in series I at the same position (3rd term) is $c + 8 = 20$.

16. Difference between wrong term of series II and correct term of series I:
    $$36 - 20 = 16$$

17. Analyze 16:
    - 16 is an even number.
    - 16 is a composite number.

**Answer:** Both Even number and Composite number (Option e)

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