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Norma-Jean [14]

3 years ago

9

HEEEEEEEEEELLLLLLLLLLP

1 answer:

antoniya [11.8K]3 years ago

3 0

The answer is B, d=8.75h
It’s 8.75h because Michael earns 8.75 per hour the key word is “per hour”

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A box contains 10 red pens and 10 black pens. Which is more likely , that you will pick a red pen or a black pen out of the box?

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Share £165 in the ratio 2:3

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6 0

3 years ago

The 4th and the last terms of an A.P. are 11 and 89 respectively. If there are 30 terms in the A.P., find the A.P. and its 23rd

Natali5045456 [20]

$\underline{\underline{\large\bf{Given:-}}}$

$\red{\leadsto}\:$ $\textsf{}$ $\sf Number \: of \:terms \: in \: A.P,n = 30$

$\red{\leadsto}\:$ $\textsf{}$ $\sf Fourth \: term ,a_4 = 11$

$\red{\leadsto}\:$ $\textsf{}$ $\sf last\:term, a_{30} = 89$

$\underline{\underline{\large\bf{To Find:-}}}$

$\orange{\leadsto}\:$ $\textsf{ }$ $\sf The \: A.P.$

$\orange{\leadsto}\:$ $\textsf{ }$ $\sf 23rd\: term, a_{23}$

$\\$

$\underline{\underline{\large\bf{Solution:-}}}\\$

The nth term of A.P is determined by the formula-

$\green{ \underline { \boxed{ \sf{a_n = a+(n-1)d}}}}$

where

$\sf a = first \:term$
$\sf a_n = nth \: term$
$\sf n = number \:of \:terms$
$\sf d = common \: difference$

Since ,

$\sf a_4 = 11$

$\longrightarrow$ $\sf a+(4-1) d= 11$

$\longrightarrow$ $\sf a+3d= 11\_\_\_(1)$

$\sf a_{30}= 89$

$\longrightarrow$ $\sf a+(30-1)d=89$

$\longrightarrow$ $\sf a+29d= 89\_\_\_(2)$

<u>Subtracting equation (1) from equation(2)</u>

$\begin{gathered}\\\implies\quad \sf a+29d-(a+3d) = 89-11 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a+29d-a-3d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a-a+29d-3d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf 26d = 78 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf d = \frac{78}{26} \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf d = 3 \\\end{gathered}$

Putting the value of d in equation (1) -

$\begin{gathered}\\\implies\quad \sf a+3(3) = 11 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a = 11-9 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a = 2 \\\end{gathered}$

First term of A.P, a = 2

Second term of A.P., $\sf a_2= 2+(2-1)\times 3$

$\quad\quad\quad\sf =2+3$

$\quad\quad\quad\sf =5$

Third term of A.P., $\sf a_3= 2+(3-1)\times 3$

$\quad\quad\quad\sf =2+6$

$\quad\quad\quad\sf =8$

$\longrightarrow$ Thus , The A.P is 2,5,8,. . . . . .

<u>Now,</u>

$\begin{gathered}\\\implies\quad \sf a_n = a+(n-1)d \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23 }= 2+(23-1)\times 3 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 2+22 \times 3 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 2+66 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf a_{23} = 68 \\\end{gathered}$

$\longrightarrow$ Thus , 23rd term is 68.

3 0

3 years ago