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kiruha [24]
4 years ago
7

Find an equation for the nth term of the arithmetic sequence. a16 = 21, a17 = -1

Mathematics
1 answer:
sergij07 [2.7K]4 years ago
3 0

Answer:nth term=a1 - 27n + 27

Step-by-step explanation:

first term =a1

common difference=d=-1-21

d=-27

Using the formula

Tn=a1 + d x (n-1)

nth term=a1 + (-27)(n-1)

nth term=a1 - 27n + 27

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Rationalise the denominator of: (√3 - √2)/(√3+√2) = ?​
marissa [1.9K]

Step-by-step explanation:

<h2><u>Given :-</u></h2>

(√3-√2)/(√3+√2)

<h2><u>To find :-</u></h2>

Rationalised form = ?

<h2><u>Solution:-</u></h2>

Given that

(√3-√2)/(√3+√2)

The denominator = √3+√2

The Rationalising factor of √3+√2 is √3-√2

On Rationalising the denominator then

=> [(√3-√2)/(√3+√2)]×[(√3-√2)/(√3-√2)]

=> [(√3-√2)(√3-√2)]×[(√3+√2)(√3-√2)]

=> (√3-√2)²/[(√3+√2)(√3-√2)]

=> (√3-√2)²/[(√3)²-(√2)²]

Since (a+b)(a-b) = a²-b²

Where , a = √3 and b = √2

=> (√3-√2)²/(3-2)

=> (√3-√2)²/1

=> (√3-√2)²

=> (√3)²-2(√3)(√2)+(√2)²

Since , (a-b)² = a²-2ab+b²

Where , a = √3 and b = √2

=> 3-2√6+2

=> 5-2√6

Hence, the denominator is rationalised.

<h2><u>Answer</u><u>:</u></h2>

Rationalised form of (√3-√2)/(√3+√2) is 5 - 2√6.

<h2><u>U</u><u>sed </u><u>formulae:</u><u>-</u></h2>
  • (a+b)(a-b) = a²-b²
  • (a-b)² = a²-2ab+b²
  • The Rationalising factor of √3+√2 is √3-√2
4 0
2 years ago
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