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
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
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
To prove that the 10th term is twice the 5th term, we need to show that t10 = 2 * t5, where t10 represents the 10th term and t5 represents the 5th term of the arithmetic sequence.
Given that the 6th term is 3 times the 2nd term, we can express this relationship mathematically as t6 = 3 * t2.
Using the formula for the nth term of an arithmetic sequence, tn = t1 + (n - 1) * d, where t1 is the first term and d is the common difference, we can set up two equations involving t6 and t2:
t6 = t1 + 5d ----- Equation (1) (since the 6th term is obtained by adding 5d to the first term)
t6 = 3t2 ----- Equation (2) (given)
From Equation (1), we can express t1 in terms of t6 and d:
t1 = t6 - 5d
Substituting this into Equation (2), we can solve for t2:
3t2 = t6
Dividing both sides by 3, we get:
t2 = t6 / 3
Knowing that t6 = t1 + 5d, we can substitute this expression into t2:
t2 = (t1 + 5d) / 3
Now, to prove that t10 = 2 * t5, let's express t10 and t5 in terms of t1 and d:
t10 = t1 + 9d ----- Equation (3) (since the 10th term is obtained by adding 9d to the first term)
t5 = t1 + 4d ----- Equation (4) (since the 5th term is obtained by adding 4d to the first term)
Multiplying Equation (4) by 2 gives:
2 * t5 = 2 * (t1 + 4d)
Expanding this equation:
2 * t5 = 2t1 + 8d
From Equation (3), we know that t10 = t1 + 9d. Substituting this into the expression for 2 * t5, we get:
2 * t5 = t10 - d
Therefore, if we can show that d = 0, then we can prove that t10 = 2 * t5.
From Equation (1), we had t6 = t1 + 5d. Substituting this expression into Equation (4), we get:
t5 = t6 - d
Since we are given that t6 = 3t2, we can rewrite this as:
t5 = 3t2 - d
Now, equating the two expressions for t5 obtained from the previous steps:
t5 = 3t2 - d
t5 = t10 - d
We can equate the right sides of these expressions:
3t2 - d = t10 - d
Subtracting t2 from both sides of the equation:
3t2 - t2 - d = t10 - t2 - d
Simplifying:
2t2 - d = t10 - t2 - d
Canceling the "d" terms:
2t2 = t10 - t2
Adding t2 to both sides of the equation:
2t2 + t2 = t10
Simplifying:
3t2 = t10
Since we know that t2 = t6 / 3, we can substitute this into the equation:
3 * (t6 / 3) = t10
Canceling the 3's:
t6 = t10
Therefore, the 10th term is the same as the 6th term.
Since the 6th term is 3 times the 2nd term, we can write:
t6 = 3t2
From our previous step, we established that t6 = t10, so we can rewrite this as:
t10 = 3t2
We also found earlier that 2 * t5 = t10 - d, and since we showed that d = 0, we can write:
2 * t5 = t10
Hence, we have proved that the 10th term is twice the 5th term, as desired.