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3 years ago

8

Please help on this one ?

2 answers:

alekssr [168]3 years ago

7 0

1 ║ 4 6 -1

4 10

----------------------------------------------

4 10 $\boxed{9}$

⇒ The Remainder is 9

xenn [34]3 years ago

6 0

When you complete the synthetic division you get 9 as the remainder so your answer is A. 9

Work in image, sorry it’s super messy

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3 years ago

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3 years ago

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3 years ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

8 0

3 years ago

–4k + 5 ≤ 21 i need help please

Allushta [10]

Answer :
inequality form: k ≥ -4
OR
interval form: [-4, ∞)

6 0

3 years ago

Read 2 more answers