if the first term of ap is 2 and the sum of first 5 terms is equal to 1/4 of the sum of the next five terms. find the sum offirst 30 terms and also fine 20th term

Let the first term of the arithmetic progression (AP) be \( a = 2 \), and let the common difference be \( d \). We're tasked with finding the sum of the first 30 terms and the 20th term, given that the sum of the first 5 terms is equal to \( \frac{1}{4} \) of the sum of the next 5 terms.

### Step 1: Sum of the first 5 terms
The sum of the first \( n \) terms of an arithmetic progression is given by:
\[
S_n = \frac{n}{2} \times \left(2a + (n-1) d\right)
\]
For \( n = 5 \), the sum of the first 5 terms is:
\[
S_5 = \frac{5}{2} \times \left(2a + (5-1)d\right) = \frac{5}{2} \times (4 + 4d)
\]
\[
S_5 = 5 \times (2 + 2d) = 10 + 10d
\]

### Step 2: Sum of the next 5 terms
The next 5 terms start from the 6th term to the 10th term. The sum of these terms is:
\[
S_6 \text{ to } S_{10} = \frac{5}{2} \times \left(2(a + 5d) + (5-1) d\right) = \frac{5}{2} \times (2a + 10d + 4d)
\]
\[
S_6 \text{ to } S_{10} = \frac{5}{2} \times (4 + 14d) = 5 \times (2 + 7d) = 10 + 35d
\]

### Step 3: Use the condition
We are given that the sum of the first 5 terms is equal to \( \frac{1}{4} \) of the sum of the next 5 terms:
\[
S_5 = \frac{1}{4} \times S_6 \text{ to } S_{10}
\]
Substituting the expressions we found earlier:
\[
10 + 10d = \frac{1}{4} \times (10 + 35d)
\]
Multiplying both sides by 4 to clear the fraction:
\[
40 + 40d = 10 + 35d
\]
Simplifying:
\[
40d - 35d = 10 - 40
\]
\[
5d = -30 \quad \Rightarrow \quad d = -6
\]

### Step 4: Find the sum of the first 30 terms
Now that we have \( d = -6 \), we can find the sum of the first 30 terms:
\[
S_{30} = \frac{30}{2} \times \left(2a + (30-1)d\right)
\]
Substitute \( a = 2 \) and \( d = -6 \):
\[
S_{30} = 15 \times \left(2 \times 2 + 29 \times (-6)\right) = 15 \times \left(4 - 174\right) = 15 \times (-170) = -2550
\]

### Step 5: Find the 20th term
The general term of an arithmetic progression is given by:
\[
T_n = a + (n-1) d
\]
For \( n = 20 \), we have:
\[
T_{20} = 2 + (20-1)(-6) = 2 + 19 \times (-6) = 2 - 114 = -112
\]

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

Would you like more details or have any further questions?

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Here are 5 related questions:
1. What is the sum of the first 15 terms of this AP?
2. How many terms are needed to get a sum of -3000 in this AP?
3. What is the common difference if the sum of the first 10 terms is 100?
4. How does the sum of an arithmetic progression change with a positive common difference?
5. What is the 50th term of this AP?

**Tip:** To quickly verify sums in an AP, always check the relationship between the terms and the total number of terms.

If the first term of AP is 2 and the sum of the first 5 terms is equal to 1/4 of the sum of the next five terms, find the sum of the first 30 terms and also find the 20th term.

Find the sum of the first 30 terms and the 20th term of an arithmetic progression where the first term is 2, and the sum of the first 5 terms is 1/4 of the sum of the next 5 terms. This problem involves key arithmetic progression formulas, suitable for students in Grades 10-12.

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