Math Problem Statement

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

Solution

Let the first term of the arithmetic progression (AP) be a=2a = 2, and let the common difference be dd. 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 14\frac{1}{4} of the sum of the next 5 terms.

Step 1: Sum of the first 5 terms

The sum of the first nn terms of an arithmetic progression is given by: Sn=n2×(2a+(n1)d)S_n = \frac{n}{2} \times \left(2a + (n-1) d\right) For n=5n = 5, the sum of the first 5 terms is: S5=52×(2a+(51)d)=52×(4+4d)S_5 = \frac{5}{2} \times \left(2a + (5-1)d\right) = \frac{5}{2} \times (4 + 4d) S5=5×(2+2d)=10+10dS_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: S6 to S10=52×(2(a+5d)+(51)d)=52×(2a+10d+4d)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) S6 to S10=52×(4+14d)=5×(2+7d)=10+35dS_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 14\frac{1}{4} of the sum of the next 5 terms: S5=14×S6 to S10S_5 = \frac{1}{4} \times S_6 \text{ to } S_{10} Substituting the expressions we found earlier: 10+10d=14×(10+35d)10 + 10d = \frac{1}{4} \times (10 + 35d) Multiplying both sides by 4 to clear the fraction: 40+40d=10+35d40 + 40d = 10 + 35d Simplifying: 40d35d=104040d - 35d = 10 - 40 5d=30d=65d = -30 \quad \Rightarrow \quad d = -6

Step 4: Find the sum of the first 30 terms

Now that we have d=6d = -6, we can find the sum of the first 30 terms: S30=302×(2a+(301)d)S_{30} = \frac{30}{2} \times \left(2a + (30-1)d\right) Substitute a=2a = 2 and d=6d = -6: S30=15×(2×2+29×(6))=15×(4174)=15×(170)=2550S_{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: Tn=a+(n1)dT_n = a + (n-1) d For n=20n = 20, we have: T20=2+(201)(6)=2+19×(6)=2114=112T_{20} = 2 + (20-1)(-6) = 2 + 19 \times (-6) = 2 - 114 = -112

Final Answers:

  • The sum of the first 30 terms is 2550-2550.
  • The 20th term is 112-112.

Would you like more details or have any further questions?


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.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Sum of Terms
General Term of AP

Formulas

Sum of the first n terms: S_n = (n/2) × [2a + (n-1)d]
General term of an arithmetic progression: T_n = a + (n-1)d

Theorems

Sum formula for an arithmetic progression
General term formula of an arithmetic progression

Suitable Grade Level

Grades 10-12