Answer:
by thinking a number we can simply get the values
10 , 20 , 30 , 40 , 50 , 60
but if we solve it by using the arithmetic sequence formula we will get
an=a1 + (n-1)d let an= unknown
a1= 10
a6=60 because if we add 4 arithmetic means then it will be 6 terms and 60 will be the 6th term
• lets solve for d
a(6)= 10 + (6-1)d
60 = 10 + 5d
60-10 = 5d
50 = 5d divide both sides by 5
10 = d
then just add their common difference
10+10= 20 -------> 1st term inserted
20+10= 30 -------> 2nd term inserted
30+10 = 40 -------> 3rd term inserted
40+10 = 50 -------> 4th term inserted
ANSWER: 20,30,40,and 50
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Answer:
by thinking a number we can simply get the values
10 , 20 , 30 , 40 , 50 , 60
but if we solve it by using the arithmetic sequence formula we will get
an=a1 + (n-1)d let an= unknown
a1= 10
a6=60 because if we add 4 arithmetic means then it will be 6 terms and 60 will be the 6th term
• lets solve for d
a(6)= 10 + (6-1)d
60 = 10 + 5d
60-10 = 5d
50 = 5d divide both sides by 5
10 = d
then just add their common difference
10+10= 20 -------> 1st term inserted
20+10= 30 -------> 2nd term inserted
30+10 = 40 -------> 3rd term inserted
40+10 = 50 -------> 4th term inserted
ANSWER: 20,30,40,and 50