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4 years ago

Find the sum of the even integers between 21 and 45

Mathematics

2 answers:

Vika [28.1K]4 years ago

6 0

22+24+26+28+30+32+34+36+38+40+42+44= 396

mario62 [17]4 years ago

4 0

Solution:

we have been asked to find the sum of the even integers between 21 and 45.

The even integers between 21 and 45 is

22, 24, 26,28, 30, 32, 34, 36,38, 40, 42,44.

its forming an arithmetic progression with

first term=22=a

number of term=12=n

common difference=24-22=2=d

The sum can be found using the formula

$S_n=\frac{n}{2}[2a+(n-1)d]\\$

Substitute the values in the above formula we get

$S_{12}=\frac{12}{2}[2*22+(12-1)2]\\ \\ S_{12}=6[44+22]\\ \\ S_{12}=6*66\\ \\S_{12}=396\\$

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4 years ago

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BaLLatris [955]

Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;

The n'th term of a Arithmetic Sequence let's say it be ${\sf T_n}$ is given by ;

${\boxed{\bf T_{n}=T_{1}+(n-1)d}}$

Where , d is the common difference

Now , here we are given with ;

${\quad \qquad \sf \blacktriangleright \blacktriangleright \blacktriangleright T_{1}=8 \: and \: T_{5}=-4}$

We have to find the 2nd , 3rd and 4th term respectively ,

Now , by using the above formula , 5th term can be written as ;

${: \implies \quad \sf T_{1}+(5-1)d=T_{5}}$

Putting the values and transposing 1st term to RHS , we have ;

${: \implies \quad \sf 4d = -4-8}$

${: \implies \quad \sf d=-\dfrac{12}{4}}$

${: \implies \quad \sf d=-3}$

Now , as we got the common difference , so we can find out the missing terms now ;

${: \implies \quad \sf T_{2}= T_{1}+(2-1)d}$

${: \implies \quad \sf T_{2}= 8 +d}$

${: \implies \quad \sf T_{2}= 8-3}$

${: \implies \quad \bf \therefore \: T_{2}= 5}$

Now

${: \implies \quad \sf T_{3}= T_{1}+(3-1)d}$

${: \implies \quad \sf T_{3}= 8 +2d}$

${: \implies \quad \sf T_{3}= 8-6}$

${: \implies \quad \bf \therefore \: T_{3}= 2}$

Also ,

${: \implies \quad \sf T_{4}= T_{1}+(4-1)d}$

${: \implies \quad \sf T_{4}= 8 +3d}$

${: \implies \quad \sf T_{4}= 8-9}$

${: \implies \quad \bf \therefore \: T_{4}= -1}$

Now , The given table can be written as ;

${\begin{array}{|c|c|c|c|c|c|}\cline{1-6} \bf n & \sf 1 & 2 & 3 & 4 & 5 \\ \cline{1-6} \bf T_{n} & \sf 8 & 5 & 2 & -1 & -4 \end{array}}$

Note :- Kindly view the answer from web , if you're not able to see the full answer from here ;

brainly.com/question/26750175

8 0

3 years ago

√₂

snow_lady [41]

Answer:

$2000 was invested at 5% and $5000 was invested at 8%.

Step-by-step explanation:

Assuming the interest is simple interest.

Simple Interest Formula

I = Prt

where:

I = interest earned.
P = principal invested.
r = interest rate (in decimal form).
t = time (in years).

Given:

Total P = $7000
P₁ = principal invested at 5%
P₂ = principal invested at 8%
Total interest = $500
r₁ = 5% = 0.05
r₂ = 8% = 0.08
t = 1 year

Create two equations from the given information:

$\textsf{Equation 1}: \quad \sf P_1+P_2=7000$

$\textsf{Equation 2}: \quad \sf P_1r_1t+P_2r_2t=I\implies 0.05P_1+0.08P_2=500$

Rewrite Equation 1 to make P₁ the subject:

$\implies \sf P_1=7000-P_2$

Substitute this into Equation 2 and solve for P₂:

$\implies \sf 0.05(7000-P_2)+0.08P_2=500$

$\implies \sf 350-0.05P_2+0.08P_2=500$

$\implies \sf 0.03P_2=150$

$\implies \sf P_2=\dfrac{150}{0.03}$

$\implies \sf P_2=5000$

Substitute the found value of P₂ into Equation 1 and solve for P₁:

$\implies \sf P_1+5000=7000$

$\implies \sf P_1=7000-5000$

$\implies \sf P_1 = 2000$

$2000 was invested at 5% and $5000 was invested at 8%.

Learn more about simple interest here:

brainly.com/question/27743947

brainly.com/question/28350785

5 0

1 year ago

Help please! If you know geometry please help!

tamaranim1 [39]

50+50=100

180-100=80

X=80

4 0

3 years ago