Question

1) Consider the arithmetic sequence 1, 8, 15, 22 · · · a) What is the common difference ? b) What will be the remainder when the terms are divided by its common difference? c) Which is the first three digit term of this sequence? d) Write the algebraic form of this sequence e) How many terms are there below 100in this sequence?

Answer 1

Answer:

D=7

Step-by-step explanation:

a. 8-1=7

d=7

C.1,8,15

Answer 2

Answer:

a) Common difference $=7$

b) The remainder when the terms are divided by its common difference will be one.

c) From $16$th term, three digit term sequence is started.

d) The algebraic form is $a_{n} = 7n-6$.

e) $15$ terms are there below $100$ in this sequence.

Step-by-step explanation:

Concept:

An arithmetic sequence is one in which each phrase grows by adding or removing a certain constant, k. In a geometric sequence, each term rises by dividing by or multiplying by a certain constant k.

Given:

An arithmetic sequence $1, 8, 15, 22.....$

To find:

We have to answer the asked questions.

Solution:

We have,

$1, 8, 15, 22....$

$a_{1} =1\\a_{2} =8\\a_{3} =15\\a_{4} =22$

a) The common difference $(d)$ is,

$d=a_{2} -a_{1}$

$8-1=7$

$d=7$

b) The remainder will be the previous term, it is shown as below:

$\frac{1}{7}$ remainder $=1$

$\frac{8}{7}$ remainder $=1$

$\frac{15}{7}$ remainder $=1$

$\frac{22}{7}$ remainder $=1$

c) $a=1, d = 7$,

$a_{n} > 99$

$a+(n-1)d > 99\\1 + (n-1)7 > 99\\1+7n-7 > 99\\7n > 99+6\\7n > 105$

$n > \frac{105}{7} \\n > 15$

Thus, $16$th term is the first three digit term of the arithmetic sequence.

d) The algebraic form of this sequence is as below:

$a_{n} = a + (n-1)d$

$= 1+(n-1)(7)\\= 1+7n-7\\= 7n-6$

$a_{n} = 7n-6$

e) As we calculated in question c), it is clear that there are $15$ terms that are below $100$.

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