Step 1: Let the first term be \(a\) and common difference be \(d\).
The \(n\)th term of A.P. is given by: \[ a_n = a + (n-1)d. \]
Step 2: Given, \[ a_{10} = a + 9d = 35, \] \[ a_{20} = a + 19d = 65. \]
Step 3: Subtracting the first equation from the second: \[ (a + 19d) - (a + 9d) = 65 - 35 \implies 10d = 30 \implies d = 3. \]
Step 4: Substitute \(d = 3\) into \(a + 9d = 35\): \[ a + 9 \times 3 = 35 \implies a + 27 = 35 \implies a = 8. \]
Step 5: Check options, \(a=8\) is option (B).
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |