a. what is the remainder on dividing the terms of arithmetic sequence 100,107,114 by 7?
b. Write the sequence of all three digit numbers which leaves the remainder 3 on division by 7?
c. Which is the last term of this sequence?
Please answer this question correctly....I will Mark you as brainlist.
Share
Answer:
a) The remainder will be 2,
First off, notice that the arithmetic sequence has a difference of 7 between consequtive terms. As the remainder of 7n7n when dividing by 7 is 0 for all n \in \mathbb{Z}n∈Z then all of the terms in the sequence have the same remainder when dividing by 7. Then finding the remainder of 100 when dividing by 7 is enough to answer the question. We then notice that 98 = 7 \times 1498=7×14 is the largest multiple of 7 that is less than 100. As 100-98 = 2, then the remainder of 100 when divided by 7 is 2. Consequently the remainder of the terms of the arithemtic sequence when divided by 7 is 2.
b) 129 number are divided by 7 which leaves remainder is 3. According to the give figure
c) Finding the Number of Terms in a Finite Arithmetic Sequence
Finding the Number of Terms in a Finite Arithmetic SequenceFind the common difference d.
Finding the Number of Terms in a Finite Arithmetic SequenceFind the common difference d.Substitute the common difference and the first term into a n = a 1 + d ( n − 1 ) \displaystyle {a}_{n}={a}_{1}+d\left(n - 1\right) an=a1+d(n−1).
Finding the Number of Terms in a Finite Arithmetic SequenceFind the common difference d.Substitute the common difference and the first term into a n = a 1 + d ( n − 1 ) \displaystyle {a}_{n}={a}_{1}+d\left(n - 1\right) an=a1+d(n−1).Substitute the last term for an and solve for n.