Answer:
Its either 300 or 315
Step-by-step explanation:
Mainly because 20 ×15= 300 but the triangles on the side, I can't figure it out. I eliminated 82 and 420, 420 mainly because I got it wrong and 82 doesn't make sense.
Answer:
a₁ = - 24
Step-by-step explanation:
The n th term of an AP is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₇ = 2a₅ , then
a₁ + 6d = 2(a₁ + 4d) = 2a₁ + 8d ( subtract 2a₁ + 8d from both sides )
- a₁ - 2d = 0 → (1)
The sum to n terms of an AP is
= 
[ 2a₁ + (n - 1)d ]
Given 
= 84 , then
(2a₁ + 6d) = 84
3.5(2a₁ + 6d) = 84 ( divide both sides by 3.5 )
2a₁ + 6d = 24 → (2)
Thus we have 2 equations
- a₁ - 2d = 0 → (1)
2a₁ + 6d = 24 → (2)
Multiplying (1) by 3 and adding to (2) will eliminate d
- 3a₁ - 6d = 0 → (3)
Add (2) and (3) term by term to eliminate d
- a₁ = 24 ( multiply both sides by - 1 )
a₁ = - 24
So,
First, we find the prime factored form (P.F.F.) of 65 and 91.
P.F.F. of 65: 5 * 13
P.F.F. of 91: 7 * 13
Find the common numbers
13
13 = G.C.F.
Note: This method works for finding any G.C.F. (find common primes)
not understanding what you are saying?
Answer:
The series is absolutely convergent.
Step-by-step explanation:
By ratio test, we find the limit as n approaches infinity of
|[a_(n+1)]/a_n|
a_n = (-1)^(n - 1).(3^n)/(2^n.n^3)
a_(n+1) = (-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)
[a_(n+1)]/a_n = [(-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)] × [(2^n.n^3)/(-1)^(n - 1).(3^n)]
= |-3n³/2(n+1)³|
= 3n³/2(n+1)³
= (3/2)[1/(1 + 1/n)³]
Now, we take the limit of (3/2)[1/(1 + 1/n)³] as n approaches infinity
= (3/2)limit of [1/(1 + 1/n)³] as n approaches infinity
= 3/2 × 1
= 3/2
The series is therefore, absolutely convergent, and the limit is 3/2
