If the sum of the first n terms of an AP is 4n - n2, what is the first term (that is S1)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.
- Sum of first n terms, Sₙ = 4n - n²
Sum of first term, S₁ = 4 × 1 - 1² = 4 - 1 = 3
Sum of first two terms, S₂ = 4 × 2 - 2² = 8 - 4 = 4
Sum of first three terms, S₃ = 4 × 3 - 3² = 12 - 9 = 3
Second term, a₂ = S₂ - S₁ = 4 - 3 = 1
Third term, a₃ = S₃ - S₂ = 3 - 4 = - 1
Tenth term, a₁₀ = S₁₀ - S₉
= (4 × 10 - 102) - (4 × 9 - 92)
= (40 - 100) - (36 - 81)
= - 60 + 45
= - 15
nth term, aₙ = Sₙ - Sₙ₋₁
= [4n - n²] - [4 (n - 1) - (n - 1)² ]
= 4n - n² - 4n + 4 + (n - 1)²
= 4 - n² + n² - 2n + 1
= 5 - 2n
If the sum of the first n terms of an AP is 4n - n², what is the first term (that is S1)? What is the sum of the first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms
Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 11
If the sum of the first n terms of an AP is 4n - n2, then the first term is equal to 3, the sum of first two terms is equal to 4, the second term is equal to 1, and 3rd term, 10th term and the nth terms are equal to -1, -15, (5 - 2n) respectively.
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