Asked by Naomi

The 6th term of an arithmetic sequence is 3 times the 2nd term. Prove that the 10th term is twice the 5th term.

tn=t1+(n-1)d
Answered by Reiny
term6 = a+ 5d
term2 = a+d

a+5d = 3(a+d)
a+5d = 3a + 3d
2d = 2a
d = a

term10 = a+9d = 10a , since d = a
term 5 = a+4d = 5d

so clearly term10 = 2 x term5
Answered by Damon
t6 = t1 + 5 d
t2 = t1 + d
so
t1 +5 d = 3 t1 + 3 d
so
2 t1 = 2 d
d = t1

t10 = t1 + 9d = 10 d

t5 = t1 + 4 d = 5 d

lo and behold 10 d/ 5 d = 2
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