Question

The algebraic form of an arithmetic sequence is 3+5n (a) Find it's common difference (b) Find its first term. (c) form the sequence ​

Answer 1

Answer:

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Step-by-step explanation:

Answer:

\begin{gathered} a1 = 3 - 5(1) \\ 3 - 5 = - 2 \\ a2 =3 - 5(2) \\ = - 7 \\ common \: difference \\ = - 7 - ( - 2) \\ = - 7 + 2 \\ = - 5 \\ sequence \\ - 2. - 7. - 13. - 18\end{gathered}a1=3−5(1)3−5=−2a2=3−5(2)=−7commondifference=−7−(−2)=−7+2=−5sequence−2.−7.−13.−18

Answer 2

Answer:

the first term is 8 and common difference is 5

Step-by-step explanation:

Hint: Use the fact that if a sequence is given by {an=f(n)}, then the first term of the sequence is determined by putting n = 1 in the formula for the general term of the sequence, i.e. a=f(1). Hence determine the first term of the arithmetic sequence. Use the fact that if a,b are any two integers, then there exist unique integers r and q such that a=bq+r,0≤r≤b−1. Here r is known as the remainder of the division of a by b. Hence determine the remainder obtained on dividing an by 5.

Complete step-by-step answer:

We know that if a sequence is given by {an=f(n)}, then the first term of the sequence is determined by putting n = 1 in the formula for the general term of the sequence, i.e. a=f(1).

Now, we have

an=5n+3

Hence, we have

a=5(1)+3=8

Hence the first term of the arithmetic sequence is 8.

Now, we have

an=5n+3 (i)

Also by Euclid’s division lemma, if a,b are any two integers, then there exist unique integers r and q such that a=bq+r,0≤r≤b−1.

Hence, we have

an=5q+r,0≤r≤4 (ii)

Since q and r are unique, from equations (i) and (ii), we get

n = q and r = 3.

Hence the remainder obtained on dividing the term of the sequence by 5 is 3.

Note: Alternative solution for part a:

We have

an=5n+3=5(n−1+1)+3=5+3+5(n−1)

Hence, we have

an=8+5(n−1)

Comparing with the formula for the general term of an arithmetic sequence, i.e. an=a+(n−1)d, we get

a=8,d=5

Hence, the first term of the sequence is 8.