Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule. A(n) = –3 + (n – 1)(–2.2)

1 answer:

7 0

A(n) = a1 + (n - 1) * d
A(n) = -3 + (n - 1) * -2.2
n = term to find
a1 = first term = -3
d = common difference = -2.2

we can already see that the first term is -3 <==

4th term..
A(4) = -3 + (4 - 1) * -2.2
A(4) = -3 + 3(-2.2)
A(4) = -3 - 6.6
A(4) = - 9.6 <== 4th term

10th term...
A(10) = -3 + (10 - 1) * -2.2
A(10) = -3 + 9 * - 2.2
A(10) = -3 - 19.8
A(10) = - 22.8 <===10th term

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What is the coefficient of the x^5y^5- term in the biomial expansion of (2x-3y)^10

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<u>Answer:</u>

The coefficient of $x^{5} \times y^{5} \text { is }=\left(252 \times 2^{5} \times(-3)^{5}\right)=252 \times 32 \times 243=1959552$

<u>Solution: </u>

The given expression is $(2 x-3 y)^{10}$

As per binomial theorem, we know,

$(x+y)^{n}=\sum n C_{k} x^{n-k} y^{k}$

Now here a = 2x, b = (- 3y) and n = 10 and k = 0,1,2,….10

Now $x^{5}\times y^5$ will be the 6 term where k =5

Now, $\mathrm{T}_{6}=10 \mathrm{C}_{5} \times(2 \mathrm{x})^{(10-5)} \times(-3 \mathrm{y})^{5}=10 \mathrm{C}_{5} 2^{5} \times \mathrm{x}^{5} \times(-3)^{5} \times \mathrm{y}^{5}$

So, the coefficient of $x^{5} \times y^{5} \text { is }=10\left(5 \times 2^{5} \times(-3)^{5}\right$ .

$10 \mathrm{C}_{5}=\frac{10 !}{5 ! \times(10-5) !}=\frac{10 !}{5 !+5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 !}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=\left(\frac{30240}{120}\right)=252$