If the first term of an A.P. is 2 and the sum of the first five terms is equal to one fourth of the sum of the next five terms, find
(i) the 20th term
(ii) the sum of first 30 terms.

Answer

To solve the problem step by step, we will follow the information given in the question and the properties of arithmetic progressions (A.P.).

### Given:
- First term of the A.P. \( a_1 = 2 \)
- Let the common difference be \( d \).

### Step 1: Find the sum of the first five terms
The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For the first five terms (\( n = 5 \)):
\[ S_5 = \frac{5}{2} \times (2 \cdot 2 + (5-1)d) = \frac{5}{2} \times (4 + 4d) = \frac{5}{2} \times 4(1 + d) = 10(1 + d) \]

### Step 2: Find the sum of the next five terms
The next five terms are \( a_6, a_7, a_8, a_9, a_{10} \). The sum of these terms can be calculated as:
\[ S_{10} - S_5 \]
Where \( S_{10} \) is the sum of the first ten terms:
\[ S_{10} = \frac{10}{2} \times (2 \cdot 2 + (10-1)d) = 5 \times (4 + 9d) = 20 + 45d \]
Thus, the sum of the next five terms is:
\[ S_6 + S_7 + S_8 + S_9 + S_{10} = S_{10} - S_5 = (20 + 45d) - (10 + 10d) = 10 + 35d \]

### Step 3: Set up the equation based on the problem statement
According to the problem, the sum of the first five terms is equal to one-fourth of the sum of the next five terms:
\[ S_5 = \frac{1}{4} S_{6 \text{ to } 10} \]
Substituting the sums we found:
\[ 10(1 + d) = \frac{1}{4}(10 + 35d) \]

### Step 4: Solve for \( d \)
Multiply both sides by 4 to eliminate the fraction:
\[ 40(1 + d) = 10 + 35d \]
Expanding gives:
\[ 40 + 40d = 10 + 35d \]
Rearranging terms:
\[ 40d - 35d = 10 - 40 \]
\[ 5d = -30 \]
\[ d = -6 \]

### Step 5: Find the 20th term
The \( n \)-th term of an A.P. is given by:
\[ a_n = a + (n-1)d \]
For the 20th term (\( n = 20 \)):
\[ a_{20} = 2 + (20-1)(-6) = 2 + 19 \cdot (-6) = 2 - 114 = -112 \]

### Step 6: Find the sum of the first 30 terms
Using the sum formula:
\[ S_{30} = \frac{30}{2} \times (2a + (30-1)d) = 15 \times (2 \cdot 2 + 29 \cdot (-6)) \]
Calculating:
\[ = 15 \times (4 - 174) = 15 \times (-170) = -2550 \]

### Final Answers:
(i) The 20th term is \( -112 \).
(ii) The sum of the first 30 terms is \( -2550 \).