Help!?? Sequence and Series question!

Find a, b and c such that {3, a, b, c, 48} is an AP and is a GP.

1 Answer
May 18, 2018

For AP #a=57/4; color(white)("d")b= 51/2; color(white)("d")c=147/4#

For GP #a=6;color(white)("d")b=12:color(white)("d")c=24#

Explanation:

#color(blue)("Consider the Arithmetic version")#

#a_1=a_1+0d = 3color(white)("d")........# where #d# is the common difference
#a_2=a_1+0d+d#
#a_3=a_1+0d+d+d#
#a_4=a_1+0d+d+d+d #
#a_5=a_1+0d+d+d+d+d -> a_1+4d#

So for any #n# we have: #color(red)(a_n=3+(n-1)d)#

Given that #a_5=3+4d=48#

#color(white)("dddd")=>color(red)(color(white)("d")d=(48-3)/4 = 45/4)#

#a_1=3 color(white)("d")larr" Given"#

#a_2=3+45/4(2-1) = 3+(45/4xx1) = 57/4#

#a_3=3+45/4(3-1) = 3+(45/4xx2) = 51/2 #

#a_4=3+45/4(4-1)=3+(45/4xx3) = 147/4#

#a_5 = 3+45/4(5-1)=3+(45/4xx4) = 48 larr" As given"#

For AP #a=57/4; color(white)("d")b= 51/2; color(white)("d")c=147/4#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the Geometric version")#

#a_1=a_1r^0=3#
#a_2=a_1r#
#a_3=a_1r^2#
#a_4=a_1r^3#
#a_5=a_1r^4=48#

General rule: #a_n=a_1r^(n-1)#

Given that #a_1=3# then #a_5=3r^4=48#

#r^4=48/3 = 16#

#r=root(4)(16) = 2#

Thus
#a_2=a=3(2)^(2-1)=6#
#a_3=b=3(2)^(3-1)=12#
#a_4=c=3(2)^(4-1)=24#

For GP #a=6;color(white)("d")b=12:color(white)("d")c=24#