Two arithmetic progression have thd same first and last terms.the first arithmetic progression has 21 terms with a common difference of 9.How many terms has the other arithmetic progression if its common difference is 4?working and answer.thans
46
Two arithmetic progression have thd same first
and last terms.the first arithmetic progression has
21 terms with a common difference of 9.How
many terms has the other arithmetic progression
if its common difference is 4?working and
answer.thans
46
To find the number of terms in the second arithmetic progression, you need to compare the common difference and the number of terms of the first progression to those of the second progression.
Given information:
- First arithmetic progression:
- Number of terms (n₁) = 21
- Common difference (d₁) = 9
- Second arithmetic progression:
- Common difference (d₂) = 4
- Number of terms (n₂) = ?
To find the number of terms in the second arithmetic progression (n₂), we can use the formula for the nth term of an arithmetic progression:
nth term = a + (n - 1)d,
where a is the first term, n is the number of terms, and d is the common difference.
Since the first and last terms are the same, we know that in both progressions:
First term (a₁) = Last term (a₂)
For the first progression:
a₁ = a₂ = a + (n₁ - 1)d₁
For the second progression:
a₁ = a₂ = a + (n₂ - 1)d₂
Equating the first and last terms of both progressions, we have:
a + (n₁ - 1)d₁ = a + (n₂ - 1)d₂
Substituting the given values:
a + (21 - 1) * 9 = a + (n₂ - 1) * 4
9n₁ = 4n₂
Dividing both sides of the equation by 4, we get:
(9n₁) / 4 = n₂
Substituting the given values:
(9 * 21) / 4 = n₂
189 / 4 ≈ 47.25
Since the number of terms cannot be a fraction, we round down to the nearest whole number.
Therefore, the second arithmetic progression has 47 terms when the common difference is 4.