Answer:
Step-by-step explanation:
The 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 a common difference and n is the number of terms and l is the last term.
(i) Given a = 5, d = 3, aₙ = 50 , find n and Sₙ.
Given,
First term, a = 5
Common difference, d = 3
nth term, l = aₙ = 50
aₙ = a + (n - 1)d
50 = 5 + (n - 1)3
45 = (n - 1)3
15 = n - 1
n = 16
Sustitute the value of n, to find sum of the n terms.
Sₙ = n/2 [a + l]
S₁₆ = 16/2 [ 5 + 50]
= 8 × 55
= 440
(ii) Given a = 7, a₁₃ = 35 , find d and S₁₃.
Given,
First term, a = 7
13th term, l = a₁₃ = 35
aₙ = a + (n - 1)d
a₁₃ = a + (13 - 1) d
35 = 7 + 12d
35 - 7 = 12d
d = 28/12
d = 7/3
Sₙ = n/2 [a + l]
S₁₃ = 13/2 [7 + 35]
= 13/2 × 42
= 13 × 21
= 273
(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
Given,
12th term, a₁₂ = 37
Common Difference, d = 3
aₙ = a + (n - 1)d
a₁₂ = a + (12 - 1) 3
37 = a + 33
a = 4
Substitute the value of a to find the sum of n terms.
Sₙ = n / 2 [a + l]
S₁₂ = 12 / 2 [4 + 37]
= 6 × 41
S₁₂ = 246
(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
Given,
3rd term, a₃ = 15
Sum up to ten terms, S₁₀ = 125.
aₙ = a + (n - 1)d
a₃ = a + (3 - 1) d
15 = a + 2d - - - - - Equation (i)
Sₙ = n/2 [2a + (n - 1) d]
S₁₀ = 10/2 [2a + (10 - 1) d]
125 = 5 [2a + 9d]
25 = 2a + 9d - - - - - Equation (ii)
On multiplying equation (i) by 2, we obtain
30 = 2a + 4d - - - - - Equation (iii)
On subtracting equation (iii) from Equation (ii), we obtain
- 5 = 5d
d = - 1
From equation (i),
15 = a + 2 (- 1)
15 = a - 2
a = 17
a₁₀ = a + (10 - 1) d
a₁₀ = 17 + 9 (- 1)
a₁₀ = 17 - 9
a₁₀ = 8
(v) Given d = 5, S₉ = 75 , find a and a₉.
Given,
Common difference, d = 5
Sum up to nine terms, S₉ = 75
Sₙ = n/2 [2a + (n - 1) d]
75 = 9/2 [2a + (9 - 1) 5]
75 = 9/2 (2a + 40)
25 = 3(a + 20)
25 = 3a + 60
a = - 35 / 3
We know that nth term of the AP is given by formula aₙ = a + (n - 1)d
a₉ = a + (9 - 1) × 5
= - 35 / 3 + 8 × 5
= - 35 / 3 + 40
= (- 35 + 120) / 3
= 85 / 3
(vi) Given a = 2, d = 8, Sₙ = 90 , find n and an.
Given,
First term, a = 2
Common difference, d = 8
Sum up to nth terms, Sₙ = 90
Sₙ = n/2 [2a + (n - 1) d]
90 = n/2 [4 + (n - 1) 8]
90 = n [2 + (n - 1)4]
90 = n [2 + 4n - 4]
90 = n [4n - 2]
90 = 4n² - 2n
4n² - 2n - 90 = 0
4n² - 20n + 18n - 90 = 0
4n (n - 5) + 18(n - 5) = 0
(n - 5)(4n + 18) = 0
Either (n - 5) = 0 or (4n + 18) = 0
n = 5 or n = - 9/2
However, n can neither be negative nor fractional.
Therefore, n = 5
aₙ = a + (n - 1) d
a₅ = 2 + (5 - 1)8
a₅ = 2 + 4 × 8
a₅ = 2 + 32
a₅ = 34
(vii) Given a = 8, aₙ = 62, Sₙ = 210 , find n and d.
Given,
First term, a = 8
nth term, l = aₙ = 62
Sum up to nth terms, Sₙ = 210
Sₙ = n / 2 [a + l]
210 = n / 2 [8 + 62]
210 = n × 35
n = 6
We know that nth term of the AP is given by aₙ = a + (n - 1)d
62 = 8 + (6 - 1) d
62 - 8 = 5d
54 = 5d
d = 54/5
(viii) Given aₙ = 4, d = 2, Sₙ = - 14 , find n and a
Given,
Common difference d = 2
nth term, l = aₙ = 4
Sum up to nth terms, Sₙ = -14
We know that nth term of AP is aₙ = a + (n - 1)d
4 = a + (n - 1)2
4 = a + 2n - 2
a = 6 - 2n .... (1)
Sₙ = n/2 [a + l]
- 14 = n/2 [6 - 2n + 4] [from(1)]
- 14 = n (5 - n)
- 14 = 5n - n²
n² - 5n - 14 = 0
n² - 7n + 2n -14 = 0
n (n - 7) + 2 (n - 7) = 0
(n - 7)(n + 2) = 0
Either n - 7 = 0 or n + 2 = 0
n = 7 or n = - 2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (1), we obtain
a = 6 - 2n
a = 6 - 2 × 7
a = 6 - 14
a = - 8
(ix) Given a = 3, n = 8, S = 192 , find d.
Given,
First term, a = 3
Number of terms, n = 8
Sum up to nth terms, Sₙ = 192
Sₙ = n/2 [2a + (n - 1) d]
192 = 8/2 [2 × 3 + (8 - 1) d]
192 = 4[6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.
Given,
Last term, l = aₙ = 28
Number of terms, n = 9
Sum up to nth terms, Sₙ = 144
Sₙ = n/2 [a + l]
144 = 9/2 (a + 28)
32 = a + 28
a = 4
