series

How many terms of the series

21+18+15+.....

must be taken to amount 66? Explain the double answer

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Without loss of generality, the sequence here in this problem is an arithmetic sequence.

You use the formula for series where S(n) = (n/2)(2 a(1) + (n-1)(d))

Using the given a(1) = 21, d = -3, and S(n) = 66,

66 = (n/2)(2(21)+(-3)(n-1))

66 = (n/2)(42-3n+3)

132 = (n)(-3n+45)

132 = -3n^2 + 45n

0 = -3n^2+45n-132

0 = 3n^2 - 45n + 132

0 = (3n - 33)(n - 4)

n = 11 or n = 4.

Therefore, when summed up to 11 or 4 terms the sum will be 66.

Explanation: Both answers are correct since when you sum up the terms in this sequence, it gives 66. Now how about the big one which is 11? This is also correct since when the sequence continues up to infinity, it will yield a negative number. For short, we have neutralizer for positive numbers added.

John Ashley Capellan - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$