How do I solve this arithmetic sequence and find the next four terms?

Mathematics

2 answers:

ololo11 [35]2 years ago

6 0

Answer:

252

Step-by-step explanation:

t50 = t1 + (n - 1)(d)

= 7 + (50 - 1) (5)

= 7 + (49)(5)

= 7 + 245

= 252

laila [671]2 years ago

3 0

Answer:

The 50ᵗʰ term of this sequence is 252

Step-by-step explanation:

Here,

First Term = a₁ = 7

Common Difference = (d) = 5

Now, For 50ᵗʰ term, n = 50

aₙ = a + (n - 1)d

a₅₀ = 7 + (50 - 1) × 5

a₅₀ = 7 + 49 × 5

a₅₀ = 7 + 245

a₅₀ = 252

Thus, The 50ᵗʰ term of this sequence is 252

-TheUnknownScientist

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Mr .Baker has baked some muffins. If he packs them in boxes of 4, he will have 3 left over. If he packs them in boxes of 5, he w

babunello [35]

Answer:

$\boxed{43}$

Step-by-step explanation:

I think the easiest way to solve this problem is by brute force: trial and error.

We must find numbers that when divided by

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Here is a list of multiples of the integer n that satisfy the three conditions. $\begin{array}{c|ccc}\mathbf{n} & \mathbf{4n+ 3} & \mathbf{5n + 3} & \mathbf{6n + 1}\\1 & 7 & 8 & 7\\2 & 11 & 13 & 13\\3 & 15 & 18 & 19\\4 & 19 & 23 & 25\\5 & 23 & 28 & 31\\6 & 27 & 33 & 37\\7 & 31 & 38 & \mathbf{43}\\8 & 35 & \mathbf{43} & 49\\9 & 39 & 48 & 55\\10 & \mathbf{43} & 53 & 61\\\end{array}\\\text{The only number that is common to all three lists is $\mathbf{43}$.}\\\text{The smallest possible number of muffins Mr. Baker could have baked is $\boxed{\mathbf{43}}$.}$