Finite arithmetic sequence

How many numbers should be inserted between the numbers 1 and 25 so that all numbers create a finite arithmetic sequence and that the sum of all members of this group is 117?

Correct result:

n =  7

Solution:

 x (1+25)/2=117  26x=234  x=9 n=x2=92=7   Correctness test:  a1=1 a9=25 d=a1+a991=1+2591=134=3.25 a2=a1+d=1+3.25=174=4.25 a3=a2+d=4.25+3.25=152=7.5 a4=a3+d=7.5+3.25=434=10.75 a5=a4+d=10.75+3.25=14 a6=a5+d=14+3.25=694=17.25 a7=a6+d=17.25+3.25=412=20.5 a8=a7+d=20.5+3.25=954=23.75 \ \\ x \cdot \ (1+25)/2=117 \ \\ \ \\ 26x=234 \ \\ \ \\ x=9 \ \\ n=x-2=9-2=7 \ \\ \ \\ \text{ Correctness test: } \ \\ a_{1}=1 \ \\ a_{9}=25 \ \\ d=\dfrac{ a_{1}+a_{9} }{ 9-1 }=\dfrac{ 1+25 }{ 9-1 }=\dfrac{ 13 }{ 4 }=3.25 \ \\ a_{2}=a_{1}+d=1+3.25=\dfrac{ 17 }{ 4 }=4.25 \ \\ a_{3}=a_{2}+d=4.25+3.25=\dfrac{ 15 }{ 2 }=7.5 \ \\ a_{4}=a_{3}+d=7.5+3.25=\dfrac{ 43 }{ 4 }=10.75 \ \\ a_{5}=a_{4}+d=10.75+3.25=14 \ \\ a_{6}=a_{5}+d=14+3.25=\dfrac{ 69 }{ 4 }=17.25 \ \\ a_{7}=a_{6}+d=17.25+3.25=\dfrac{ 41 }{ 2 }=20.5 \ \\ a_{8}=a_{7}+d=20.5+3.25=\dfrac{ 95 }{ 4 }=23.75



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