QuestionRank the sums $\mathrm{S}_{30}$ of these series from least to greatest: $4+15+26+\ldots$ , $77+79+81+\ldots$ , $12+17+22+\ldots$ , $-178-154-130-\ldots$ .

Studdy Solution

STEP 1

Assumptions1. Each of the given series is an arithmetic series. . We need to find the sum of the first30 terms for each series, denoted as $_{30}$ .
3. We need to rank these sums from least to greatest.

STEP 2

The sum of the first n terms of an arithmetic series can be calculated using the formula $_n = \frac{n}{2} (a + l)$ where $n$ is the number of terms, $a$ is the first term, and $l$ is the last term.

STEP 3

Since we are given that the series are arithmetic, the last term $l$ can be calculated using the formula $l = a + (n -1)d$ where $d$ is the common difference.

STEP 4

Calculate the common difference for the first series $d =15 -4 =11$

STEP 5

Calculate the last term for the first series $l =4 + (30 -1) \times11 =325$

STEP 6

Calculate $_{30}$ for the first series $_{30} = \frac{30}{2} (4 +325) =4945$

STEP 7

Repeat steps4 to6 for the second series. The common difference is $d =79 -77 =2$ The last term is $l =77 + (30 -1) \times2 =135$ And $_{30}$ is $_{30} = \frac{30}{2} (77 +135) =3180$

STEP 8

Repeat steps4 to6 for the third series. The common difference is $d =17 -12 =5$ The last term is $l =12 + (30 -1) \times5 =157$ And $_{30}$ is $_{30} = \frac{30}{2} (12 +157) =2535$

STEP 9

Repeat steps4 to6 for the fourth series. The common difference is $d = -154 - (-178) =24$ The last term is $l = -178 + (30 -) \times24 =494$ And $_{30}$ is $_{30} = \frac{30}{2} (-178 +494) =474$

STEP 10

Now that we have the sums for each series, we can rank them from least to greatest. The ranking is. $12+17+22+27+\ldots$ ( $_{30} =2535$ )
2. $77+79+81+83+\ldots$ ( $_{30} =3180$ )
3. $-178-154-130-106-\ldots$ ( $_{30} =4740$ )
4. $4+15+26+37+\ldots$ ( $_{30} =4945$ )

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QuestionRank the sums $\mathrm{S}_{30}$ of these series from least to greatest: $4+15+26+\ldots$ , $77+79+81+\ldots$ , $12+17+22+\ldots$ , $-178-154-130-\ldots$ .

Studdy Solution

STEP 1

Assumptions1. Each of the given series is an arithmetic series. . We need to find the sum of the first30 terms for each series, denoted as $_{30}$ .
3. We need to rank these sums from least to greatest.

STEP 2

The sum of the first n terms of an arithmetic series can be calculated using the formula $_n = \frac{n}{2} (a + l)$ where $n$ is the number of terms, $a$ is the first term, and $l$ is the last term.

STEP 3

Since we are given that the series are arithmetic, the last term $l$ can be calculated using the formula $l = a + (n -1)d$ where $d$ is the common difference.

STEP 4

Calculate the common difference for the first series $d =15 -4 =11$

STEP 5

Calculate the last term for the first series $l =4 + (30 -1) \times11 =325$

STEP 6

Calculate $_{30}$ for the first series $_{30} = \frac{30}{2} (4 +325) =4945$

STEP 7

Repeat steps4 to6 for the second series. The common difference is $d =79 -77 =2$ The last term is $l =77 + (30 -1) \times2 =135$ And $_{30}$ is $_{30} = \frac{30}{2} (77 +135) =3180$

STEP 8

Repeat steps4 to6 for the third series. The common difference is $d =17 -12 =5$ The last term is $l =12 + (30 -1) \times5 =157$ And $_{30}$ is $_{30} = \frac{30}{2} (12 +157) =2535$

STEP 9

Repeat steps4 to6 for the fourth series. The common difference is $d = -154 - (-178) =24$ The last term is $l = -178 + (30 -) \times24 =494$ And $_{30}$ is $_{30} = \frac{30}{2} (-178 +494) =474$

STEP 10

Now that we have the sums for each series, we can rank them from least to greatest. The ranking is. $12+17+22+27+\ldots$ ( $_{30} =2535$ )
2. $77+79+81+83+\ldots$ ( $_{30} =3180$ )
3. $-178-154-130-106-\ldots$ ( $_{30} =4740$ )
4. $4+15+26+37+\ldots$ ( $_{30} =4945$ )

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QuestionRank the sums S30\mathrm{S}_{30}S30​ of these series from least to greatest: 4+15+26+…4+15+26+\ldots4+15+26+…, 77+79+81+…77+79+81+\ldots77+79+81+…, 12+17+22+…12+17+22+\ldots12+17+22+…, −178−154−130−…-178-154-130-\ldots−178−154−130−….

Studdy Solution

STEP 1

Assumptions1. Each of the given series is an arithmetic series. . We need to find the sum of the first30 terms for each series, denoted as 30_{30}30​.3. We need to rank these sums from least to greatest.

STEP 2

The sum of the first n terms of an arithmetic series can be calculated using the formulan=n2(a+l)_n = \frac{n}{2} (a + l)n​=2n​(a+l)where nnn is the number of terms, aaa is the first term, and lll is the last term.

STEP 3

Since we are given that the series are arithmetic, the last term lll can be calculated using the formulal=a+(n−1)dl = a + (n -1)dl=a+(n−1)dwhere ddd is the common difference.

STEP 4

Calculate the common difference for the first seriesd=15−4=11d =15 -4 =11d=15−4=11

STEP 5

Calculate the last term for the first seriesl=4+(30−1)×11=325l =4 + (30 -1) \times11 =325l=4+(30−1)×11=325

STEP 6

Calculate 30_{30}30​ for the first series30=302(4+325)=4945_{30} = \frac{30}{2} (4 +325) =494530​=230​(4+325)=4945

STEP 7

Repeat steps4 to6 for the second series. The common difference isd=79−77=2d =79 -77 =2d=79−77=2The last term isl=77+(30−1)×2=135l =77 + (30 -1) \times2 =135l=77+(30−1)×2=135And 30_{30}30​ is30=302(77+135)=3180_{30} = \frac{30}{2} (77 +135) =318030​=230​(77+135)=3180

STEP 8

Repeat steps4 to6 for the third series. The common difference isd=17−12=5d =17 -12 =5d=17−12=5The last term isl=12+(30−1)×5=157l =12 + (30 -1) \times5 =157l=12+(30−1)×5=157And 30_{30}30​ is30=302(12+157)=2535_{30} = \frac{30}{2} (12 +157) =253530​=230​(12+157)=2535

STEP 9

STEP 10

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QuestionRank the sums $\mathrm{S}_{30}$ of these series from least to greatest: $4+15+26+\ldots$ , $77+79+81+\ldots$ , $12+17+22+\ldots$ , $-178-154-130-\ldots$ .

Assumptions1. Each of the given series is an arithmetic series. . We need to find the sum of the first30 terms for each series, denoted as $_{30}$ .
3. We need to rank these sums from least to greatest.

The sum of the first n terms of an arithmetic series can be calculated using the formula $_n = \frac{n}{2} (a + l)$ where $n$ is the number of terms, $a$ is the first term, and $l$ is the last term.

Since we are given that the series are arithmetic, the last term $l$ can be calculated using the formula $l = a + (n -1)d$ where $d$ is the common difference.

Calculate the common difference for the first series $d =15 -4 =11$

Calculate the last term for the first series $l =4 + (30 -1) \times11 =325$

Calculate $_{30}$ for the first series $_{30} = \frac{30}{2} (4 +325) =4945$

Repeat steps4 to6 for the second series. The common difference is $d =79 -77 =2$ The last term is $l =77 + (30 -1) \times2 =135$ And $_{30}$ is $_{30} = \frac{30}{2} (77 +135) =3180$

Repeat steps4 to6 for the third series. The common difference is $d =17 -12 =5$ The last term is $l =12 + (30 -1) \times5 =157$ And $_{30}$ is $_{30} = \frac{30}{2} (12 +157) =2535$