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What is the sum of the first 21 terms of the series given below?1 – 5 + 4 – 9 + 7 – 13 + 10 – 17 + ……………

Question

What is the sum of the first 21 terms of the series given below?

1 – 5 + 4 – 9 + 7 – 13 + 10 – 17 + ……………

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Solution

1. Break Down the Problem

We need to identify the pattern in the given series and compute the sum of the first 21 terms.

The series is as follows:

  • The first term is 11
  • The second term is 5-5
  • The third term is 44
  • The fourth term is 9-9
  • The fifth term is 77
  • The sixth term is 13-13
  • The seventh term is 1010
  • The eighth term is 17-17

2. Relevant Concepts

To solve this, we can identify the pattern for the terms:

  • Positive terms: 1,4,7,10,1, 4, 7, 10, \ldots
  • Negative terms: 5,9,13,17,-5, -9, -13, -17, \ldots

Positive terms: This is an arithmetic series where:

  • First term a1=1 a_1 = 1
  • Common difference d=3 d = 3

The n n -th positive term can be formulated as: Pn=1+3(n1)=3n2 P_n = 1 + 3(n-1) = 3n - 2

Negative terms: This is also an arithmetic series where:

  • First term a1=5 a_1 = -5
  • Common difference d=4 d = -4

The n n -th negative term can be formulated as: Nn=54(n1)=4n1 N_n = -5 - 4(n-1) = -4n - 1

3. Analysis and Detail

The sequence of terms alternates between positive and negative. Thus, we can express the total sum of the first 21 terms as follows:

  • For the first 21 terms:
    • There are 11 positive terms (since (21+1)/211 (21 + 1) / 2 \to 11 )
    • There are 10 negative terms (since 21/210 21 / 2 \to 10 )

Sum of positive terms SP S_P : SP=P1+P2++P11 S_P = P_1 + P_2 + \ldots + P_{11} =(3×12)+(3×22)++(3×112) = (3 \times 1 - 2) + (3 \times 2 - 2) + \ldots + (3 \times 11 - 2) =(1)+(4)+(7)+...+(31) = (1) + (4) + (7) + ... + (31) This is the arithmetic series with:

  • First term a=1 a = 1
  • Last term l=31 l = 31
  • Number of terms n=11 n = 11

The sum SP S_P is calculated as: SP=n2(a+l) S_P = \frac{n}{2} (a + l) =112(1+31)=112×32=11×16=176 = \frac{11}{2} (1 + 31) = \frac{11}{2} \times 32 = 11 \times 16 = 176

Sum of negative terms SN S_N : SN=N1+N2++N10 S_N = N_1 + N_2 + \ldots + N_{10} =(5)+(9)+(13)+...+(41) = (-5) + (-9) + (-13) + ... + (-41) This is also an arithmetic series with:

  • First term a=5 a = -5
  • Last term l=41 l = -41
  • Number of terms n=10 n = 10

The sum SN S_N is calculated as: SN=n2(a+l) S_N = \frac{n}{2} (a + l) =102(541)=5×(46)=230 = \frac{10}{2} (-5 - 41) = 5 \times (-46) = -230

4. Verify and Summarize

Now we can find the complete sum of the first 21 terms: Total Sum=SP+SN=176230=54 \text{Total Sum} = S_P + S_N = 176 - 230 = -54

Final Answer

The sum of the first 21 terms of the series is 54-54.

This problem has been solved

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