(i) How many terms of the sequence 18, 16, 14, ... should be taken so that their sum is zero?
(ii) How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?
(iii) How many terms of the A.P. 9, 17, 25,... must be taken so that their sum is 636?
(iv) How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?
Answers
Answer with Step-by-step explanation:
(i) Given :
18, 16 ,14 ,....... & Sn = 0
Here, a = 18 , d = 16 - 18 = - 2
By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]
0 = n/2 [2(18) + (n – 1) (-2)]
0 = n/2 [36 - 2n + 2]
0 = n/2[38 - 2n]
0 = 38n - 2n²
38n – 2n² = 0
2n(19 - n) = 0
19 - n = 0
n = 19
2n = 0
n = 0 (which is not possible)
Hence, number of terms is 19.
(ii)
Given :
a (first term) = - 14 , a5 = 2 and Sn = 40
By using the formula ,an = a + (n - 1)d
a5 = - 14 + (5 -1) d
2 = -14 + 4d
2 + 14 = 4d
16 = 4d
d = 16/4
d = 4
By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]
40 = n/2 [2(-14) + (n -1) (4)]
40 = n/2 [- 28 + 4n - 4]
40 = n/2 [- 32 + 4n]
40 = n/2 × 2 [- 16 + 2n]
40 = n[- 16 + 2n]
40 = 2n²- 16n
2n² – 16n - 40 = 0
2(n² - 8n - 20) = 0
n² - 8n - 20 = 0
n² – 10n + 2n – 20 = 0
[By middle term splitting]
n(n - 10) + 2(n - 10) = 0
(n – 10) (n + 2) = 0
(n – 10) or (n + 2) = 0
n = 10 and n = -2 (terms can't be negative)
Hence , the number of terms in the AP are 10.
(iii) Given :
9,17, 25 ……. & Sn = 636
Here, a = 9 d = 17 - 9 = 8
By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]
636 = n/2 [2(9) + (n – 1) 8
636 = n/2 [18 - 8n - 8]
636 = n/2 [10 - 8n]
636 = n/2 × 2 [5 - 4n]
636 = n [5 - 4n]
636 = 5n + 4n²
4n² + 5n – 636 = 0
4n² - 48n + 53n - 636 = 0
[By middle term splitting]
4n(n -12) + 53(n - 12) = 0
(4n + 53)(n - 12) = 0
(4n + 53) or (n - 12) = 0
n = 12 or n = - 53/4(terms can't be negative)
Hence, the number of terms in the AP are 12.
(iv)
Given :
A.P are 63, 60 ,57, ……
Here, a = 63 , d = 60 - 63 = - 3 & Sn = 693
By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]
693 = n/2 [2(63) + (n – 1) (-3)]
693 = n/2 [126 - 3n + 3]
693 = n/2 [129 - 3n]
693 = n/2 × 3 [43 - n]
693 × ⅔ = n [43 - n]
231 × 2 = n [43 - n]
462 = n [43 – n]
n² – 43n + 462 = 0
n² - 22n - 21n + 462 = 0
[By middle term splitting]
n(n - 22) - 21(n - 22) = 0
(n - 21) (n - 22) = 0
(n - 21) or (n - 22) = 0
n = 22 or n = 21
Hence, the number of terms in the AP are 21 & 22.
HOPE THIS ANSWER WILL HELP YOU….
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