How do I find the nth term of the arithmetic sequence whose initial term and common difference d are given below? (A) #a_1=4, d=3# (B) #a_1= -8, d=5#

2 Answers
Jun 27, 2018

(A)#" "n ^(th)=3n+1#

(B)#" "n^(th)=5n-13#

Explanation:

the #n^(th) #term of an Ap is given by

#n^(th)=a+(n-1)d#

#(A)#

#n^(th)=4+(n-1)3#

#n^(th)=4+3n-3#

#:.n ^(th)=3n+1#

#(B)

#n^(th)=-8+(n-1)5#

#n^(th)=-8+5n-5#

#:.n^(th)=5n-13#

Jun 27, 2018

a) #color(white)("d")a_n=+4+3(n-1)#

b) #color(white)("d")a_n=-8+5(n-1)#

Explanation:

#color(blue)("Solution to part A ( in detail )")#

Given: #a_1=4 and d=3#

I am doing it this way to get you used to the way sequences are written. I will be building up the the #n^("th")# term

Let the position count in the sequence be #i#

#i=1->a_i->color(white)("dd")a_1=4+ 0d larr" First term"#
#i=2->a_i->color(white)("dd")a_2=4+ 1d#
#i=3->a_i->color(white)("dd")a_3=4+ 2d#
#i=4->a_i->color(white)("dd")a_4=4+ 3d#
#i=5->a_i->color(white)("dd")a_5=4+ 4d#

Looking at the behaviour we have:

Set #d=3 and i=n# giving:

#a_i=a_1+(i-1)d#

#a_n=4+(n-1)3#

Writing this as required by convention:

#a_n=4+3(n-1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solution to part B - using the above")#

Set #a_1=-8, d=5 and i=n# giving:

#a_i=a_1+(i-1)d#

#a_n=-8+(n-1)5#

#a_n=-8+5(n-1)#