Show that a₁, a₂,... , aₙ, ... form an AP where aₙ is defined as below :
(i) aₙ = 3 + 4n (ii) aₙ = 9 - 5n
Also, find the sum of the first 15 terms in each case.
Solution:
A sequence that has common difference between any two of its consecutive terms is an arithmetic progression.
Sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d] or Sₙ = n/2 [a + l], and the nth term of an AP is aₙ = a + (n - 1)d
Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.
(i) aₙ = 3 + 4n
Given,
- nth term, aₙ = 3 + 4n
a₁ = 3 + 4 × 1 = 7
a₂ = 3 + 4 × 2 = 3 + 8 = 11
a₃ = 3 + 4 × 3 = 3 + 12 = 15
a₄ = 3 + 4 × 4 = 3 + 16 = 19
It can be observed that
a₂ - a₁ =11 - 7 = 4
a₃ - a₂ =15 - 11 = 4
a₄ - a₃ =19 - 15 = 4
So, the difference of aₙ and aₙ₋₁ is constant.
Therefore, this is an AP with common difference as 4 and first term as 7.
The sum of n terms of AP is given by thre formula Sₙ = n/2 [2a + (n - 1) d]
Sum of 15 terms,
S₁₅ = 15/2 [2 × 7 + (15 - 1) 4]
= 15/2 [14 + 14 × 4]
= 15/2 × 70
= 15 × 35
= 525
(ii) aₙ = 9 - 5n
Given,
- nth term is aₙ = 9 - 5n
a₁ = 9 - 5 × 1 = 9 - 5 = 4
a₂ = 9 - 5 × 2 = 9 - 10 = - 1
a₃ = 9 - 5 × 3 = 9 - 15 = - 6
a₄ = 9 - 5 × 4 = 9 - 20 = - 11
It can be observed that
a₂ - a₁ = (-1 ) - 4 = - 5
a₃ - a₂ = (-6) -(-1) = - 5
a₄ - a₃ = (- 11) - (- 6) = - 5
So, the difference of aₙ and aₙ₋₁ is constant.
Therefore, this is an A.P. with common difference - 5 and first term as 4.
The sum of n terms of AP is given by thre formula Sₙ = n/2 [2a + (n - 1) d]
S₁₅ = 15/2 [2 × 4 + (15 - 1)(- 5)]
= 15/2 [8 + 14 (- 5)]
= 15/2 [8 - 70]
= 15/2 × (- 62)
= - 465
☛ Check: NCERT Solutions for Class 10 Maths Chapter 5
Video Solution:
Show that a₁, a₂,... , aₙ , ...form an AP where aₙ is defined as below (i) aₙ = 3 + 4n (ii) aₙ = 9 - 5n. Also find the sum of the first 15 terms in each case
Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 10
Summary:
We have seen that a₁, a₂,... , aₙ , ...form an AP where aₙ is defined as below (i) aₙ = 3 + 4n (ii) aₙ = 9 - 5n. Also the sum of the first 15 terms in each case is 525 and - 465 respectively.
☛ Related Questions:
- 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.
- Find the sum of the first 40 positive integers divisible by 6.
- Find the sum of the first 15 multiples of 8.
- Find the sum of the odd numbers between 0 and 50.
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