Given A.P.: -11, -8, -5, ..., 49 Common difference d = -8 - (-11) = 3 Last term aₙ = 49 Using first term a₁ = -11 and d = 3, find n: aₙ = a₁ + (n-1)d 49 = -11 + (n-1)3 49 + 11 = 3(n-1) 60 = 3(n-1) 20 = n-1 n = 21 Therefore total terms = 21 4th term from end means: Position from beginning = n - 3 = 2Read more
Given A.P.: -11, -8, -5, …, 49
Common difference d = -8 – (-11) = 3
Last term aₙ = 49
Using first term a₁ = -11 and d = 3, find n:
aₙ = a₁ + (n-1)d
49 = -11 + (n-1)3
49 + 11 = 3(n-1)
60 = 3(n-1)
20 = n-1
n = 21
Therefore total terms = 21
4th term from end means:
Position from beginning = n – 3 = 21 – 3 = 18th term
Using arithmetic sequence formula:
a₁₈ = a₁ + (18-1)d
= -11 + (17)3
= -11 + 51
= 40
Hence, 40 is the correct answer.
The 4th term from the end of the A.P. -11, -8, -5, …, 49 is
Given A.P.: -11, -8, -5, ..., 49 Common difference d = -8 - (-11) = 3 Last term aₙ = 49 Using first term a₁ = -11 and d = 3, find n: aₙ = a₁ + (n-1)d 49 = -11 + (n-1)3 49 + 11 = 3(n-1) 60 = 3(n-1) 20 = n-1 n = 21 Therefore total terms = 21 4th term from end means: Position from beginning = n - 3 = 2Read more
Given A.P.: -11, -8, -5, …, 49
Common difference d = -8 – (-11) = 3
Last term aₙ = 49
Using first term a₁ = -11 and d = 3, find n:
aₙ = a₁ + (n-1)d
49 = -11 + (n-1)3
49 + 11 = 3(n-1)
60 = 3(n-1)
20 = n-1
n = 21
Therefore total terms = 21
4th term from end means:
Position from beginning = n – 3 = 21 – 3 = 18th term
Using arithmetic sequence formula:
a₁₈ = a₁ + (18-1)d
= -11 + (17)3
= -11 + 51
= 40
Hence, 40 is the correct answer.
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If 5 times the 5ᵗʰ term of an AP is equal to 8 times the 8ᵗʰ term then find its 13ᵗʰ term
Given: 5a₅ = 8a₈ where aₙ = a₁ + (n-1)d Let's substitute terms: 5[a₁ + 4d] = 8[a₁ + 7d] 5a₁ + 20d = 8a₁ + 56d 5a₁ - 8a₁ = 56d - 20d -3a₁ = 36d a₁ = -12d Now, for 13th term: a₁₃ = a₁ + 12d = -12d + 12d = 0 Thus, the 13th term of this AP is 0. Click here for more: https://www.tiwariacademy.in/ncert-soRead more
Given: 5a₅ = 8a₈
where aₙ = a₁ + (n-1)d
Let’s substitute terms:
5[a₁ + 4d] = 8[a₁ + 7d]
5a₁ + 20d = 8a₁ + 56d
5a₁ – 8a₁ = 56d – 20d
-3a₁ = 36d
a₁ = -12d
Now, for 13th term:
a₁₃ = a₁ + 12d
= -12d + 12d
= 0
Thus, the 13th term of this AP is 0.
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