The sum of the first two terms of an arimethic sequence is 15 and the sum of the next two terms is 43. Write the first 4 terms of the sequence?

2 Answers

the terms are 4, 11, 18, 25

Explanation:

The first equation is: #" "n + n + x = 15 #

The second equation is: #" " n + 2x + n + 3x = 43. #

#n# is the first term #x# is the variable of the sequence. so
#n + x" "# is the second term
#n + 2x" "# is the third term
#n + 3x" "# is the fourth term.

Solving for x in terms of

# n + 2x + n + 3x = 43 " "# adding like terms.

# 2n + 5x = 43" " # subtracting #5x# from both side sides gives

# 2n + 5x -5x = 43 -5x" " # resulting in

# 2n = 43 -5x" " # this can be substituted into the first equation.

# n + n + x = 15" "# which is the same as

# 2n + x = 15" " # now substitute # 2n = 43 -5x# into the question.

# 43 -5x + x = 15" " # combine like terms

# 43 - 4x = 15# add #4x# to both sides and subtract #15# from both sides

# 43 -15 - 4x + 4x = 15-15 + 4x" "# gives.

# 28 = 4x " "# Divide both sides by #4# to solve for #x #

# 28/4 = (4x)/4 # gives

# 7 = x " "# This is the variable of the sequence.

Putting #7# in for #x# to solve for #n# the first number in the sequence.

#n + n + 7 = 15" "# combine like terms

# 2n + 7 = 15" "# subtract 7 from both sides

# 2n + 7 -7 = 15 -7 " "# which gives

# 2n = 8 " "# Divide both sides by 2

# (2n)/2 = 8/2 # gives

#n = 4# so #4# is the first number of sequence add #7#

# n_2 = 11" " 4 + 7 = 11#

#n_3 = 18" " 11 +7 = 18#

#n_4 = 25" "18 + 7 = 25#

Jun 5, 2017

#4" "11" "18" "25#

Explanation:

The terms in an arithmetic sequence can be written in terms of the first term, #a# and the common difference, #d#.

#d# is added to each term to get to the next.

Terms #(n)#: #" "1color(white)(wwwww)2color(white)(wwwwww)3color(white)(wwwwww)4#

Values:#(T_n)" "color(red)(a)color(white)(www)color(red)((a+d))color(white)(ww)color(blue)((a+2d))color(white)(ww)color(blue)((a+3d)))#

We know that the sum of #color(red)(T_1 and T_2 = 15#
We know that the sum of #color(blue)(T_ 3 and T_4 = 43)#

#:.color(red)( a+a+d = 15)color(white)(wwww.w)rarr color(red)(2a +d = 15)...........A#
#:. color(blue)(a+2d + a +3d = 43)" "rarr color(blue)(2a +5d = 43)..........B#

The difference between A and B will give us a value for #4d#

#B - A:color(white)(wwwww) 4d=28#
#color(white)(wwwwwnwwnww) d=7#

We now know that the terms differ by #7# each time.

#:.color(red)( a+a+7 = 15)#
#:.color(red)(2a = 8)#
#:.color(red)(a=4#

Now we can write the first #4# terms:

#4" "11" "18" "25#

Let's check by adding:

#color(red)(4+11 = 15)" and "color(blue)(18+25 =43)#