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Find the nth term of an Arithmetic progression

"}, "advice": "

If the difference between successive pairs of terms is a constant then the series under examination is an arithmetic progression.

\n

Ths first term is \\(a\\) and the common difference is \\(d\\).

\n

The formula for the nth term of the series is given by:    \\(T_n=a+(n-1)d\\)

\n

In this example \\(a=\\var{a}\\),  \\(d = \\var{d}\\)  and  \\(n = \\var{n}\\)

\n

\\(T_\\var{n}=\\var{a}+\\simplify{{n}-1}*\\var{d}\\)

\n

\\(T_\\var{n}=\\var{a}+\\simplify{({n}-1)*{d}}\\)

\n

\\(T_\\var{n}=\\simplify{{a}+({n}-1)*{d}}\\)

", "variable_groups": [], "parts": [{"scripts": {}, "showFeedbackIcon": true, "prompt": "

Calculate the \\(\\var{n}th\\) term of the series.

\n

\\(T_\\var{n}=\\) [[0]]

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The first three terms of a series are given by:  

\n

\\(\\var{a} + \\simplify{{a}+{d}} + \\simplify{{a}+2*{d}}\\,+ \\, ...........\\)

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Find the sum of the first n terms of an arithmetic progression

"}, "advice": "

If the difference between successive pairs of terms is a constant then the series under examination is an arithmetic progression.

\n

Ths first term is \\(a\\) and the common difference is \\(d\\).

\n

The formula for the nth term of the series is given by:    \\(S_n=\\frac{n}{2}\\left(2a+(n-1)d\\right)\\)

\n

In this example \\(a=\\var{a}\\),  \\(d = \\var{d}\\)  and  \\(n = \\var{n}\\)

\n

\\(S_\\var{n}=\\frac{\\var{n}}{2}\\left(2*\\var{a}+(\\var{n}-1)\\var{d}\\right)\\)

\n

\\(S_\\var{n}=\\simplify{{n}/{2}}\\left(\\simplify{2{a}}+\\simplify{({n}-1)*{d}}\\right)\\)

\n

\\(S_\\var{n}=\\simplify{{n}/{2}}\\left(\\simplify{2{a}+({n}-1)*{d}}\\right)\\)

\n

\\(S_\\var{n}=\\simplify{{n}*{a}+{n}*({n}-1)*{d}/2}\\)

\n

", "tags": [], "functions": {}, "statement": "

The first three terms of a series are given by:  

\n

\\(\\var{a} + \\simplify{{a}+{d}} + \\simplify{{a}+2*{d}}\\,+ \\, ...........\\)

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Calculate the sum of the first \\(\\var{n}\\) terms of this series.

\n

\\(S_\\var{n}=\\) [[0]]

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The sum of the first \\(\\var{n1}\\) terms of an arithmetic progression is \\(\\var{s1}\\) and the \\(\\var{n2}th\\) term of the same series is \\(\\var{T}\\).

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Recall the formula for the sum of the first n terms of an arithmetic progression is \\(S_n=\\frac{n}{2}(2a+(n-1)d)\\).

\n

The sum of the first \\(\\var{n1}\\) terms of an arithmetic progression is \\(\\var{s1}\\)

\n

\\(\\frac{\\var{n1}}{2}(2a+\\simplify{{n1}-1}d)=\\var{s1}\\)                               equation (i)

\n

The formula for nth term of an arithmetic progression is \\(T_n=a+(n-1)d\\).

\n

The \\(\\var{n2}th\\) term of the same series is \\(\\var{T}\\)

\n

\\(a+\\simplify{{n2}-1}d=\\var{T}\\)                                                   equation (ii)

\n

Here we have two simultaneous equations. We can eliminate the \\(a\\) term.

\n

\\(\\var{n1}a+\\simplify{({n1}-1)*{n1}/2}d=\\var{s1}\\)                    equation (i)

\n

\\(\\var{n1}a+\\simplify{{n1}*({n2}-1)}d=\\simplify{{n1}*{T}}\\)                  equation (ii)*\\(\\var{n1}\\)

\n

\\(\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}d=\\simplify{{s1}-{n1}*{T}}\\)

\n

\\(d=\\frac{\\simplify{{s1}-{n1}*{T}}}{\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}}\\)

\n

\\(d=\\simplify{({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)

\n

Using this result and equation (ii) we can find the value for \\(a\\)

\n

\\(a+\\simplify{({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}=\\var{T}\\)

\n

\\(a=\\var{T}-\\simplify{({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)

\n

\\(a=\\simplify{{T}-({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)

\n

", "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Solving arithmetic progressions using simultaneous equations

"}, "ungrouped_variables": ["n1", "s1", "n2", "T", "d", "a"], "parts": [{"prompt": "

Calculate the value of the common difference.   \\(d\\) = [[0]]

\n

Calculate the value of the first term of the series.  \\(a\\) = [[1]]

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The sum of the first \\(\\var{n1}\\) terms of an arithmetic progression is \\(\\var{s1}\\) and the sum of the first \\(\\var{n2}\\) terms of an arithmetic progression is \\(\\var{s2}\\)

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Solving arithmetic progressions using simultaneous equations

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "parts": [{"marks": 0, "type": "gapfill", "showCorrectAnswer": true, "scripts": {}, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "variableReplacements": [], "prompt": "

Calculate the value of the common difference.   \\(d\\) = [[0]]

\n

Calculate the value of the first term of the series.  \\(a\\) = [[1]]

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Recall the formula for the sum of the first n terms of an arithmetic progression is \\(S_n=\\frac{n}{2}(2a+(n-1)d)\\).

\n

The sum of the first \\(\\var{n1}\\) terms of an arithmetic progression is \\(\\var{s1}\\)

\n

\\(\\frac{\\var{n1}}{2}(2a+\\simplify{{n1}-1}d)=\\var{s1}\\)                               

\n

\\(\\var{n1}a+\\simplify{{n1}*({n1}-1)/2}d=\\var{s1}\\)                                       equation (i)

\n

The sum of the first \\(\\var{n2}\\) terms of an arithmetic progression is \\(\\var{s2}\\)

\n

\\(\\frac{\\var{n2}}{2}(2a+\\simplify{{n2}-1}d)=\\var{s2}\\)                               

\n

\\(\\var{n2}a+\\simplify{{n2}*({n2}-1)/2}d=\\var{s2}\\)                                      equation (ii)

\n

We can eliminate the \\(a\\) term.

\n

\\(\\simplify{{n2}*{n1}}a+\\simplify{{n2}*{n1}*({n1}-1)/2}d=\\simplify{{n2}*{s1}}\\)                               equation (i) * \\(\\var{n2}\\)

\n

\\(\\simplify{{n2}*{n1}}a+\\simplify{{n1}*{n2}*({n2}-1)/2}d=\\simplify{{n1}*{s2}}\\)                               equation (ii) * \\(\\var{n1}\\)

\n

Subtracting gives

\n

\\(\\simplify{{n2}*{n1}*({n1}-{n2})/2}d=\\simplify{{n2}*{s1}-{n1}*{s2}}\\)

\n

\\(d=\\frac{\\simplify{{n2}*{s1}-{n1}*{s2}}}{\\simplify{{n2}*{n1}*({n1}-{n2})/2}}\\)

\n

\\(d=\\simplify{2*({n2}*{s1}-{n1}*{s2})/({n2}*{n1}*({n1}-{n2}))}\\)

\n

Inserting this value in for \\(d\\) in equation (i) gives

\n

\\(\\var{n1}a+\\simplify{{n1}*({n1}-1)/2}(\\simplify{2*({n2}*{s1}-{n1}*{s2})/({n2}*{n1}*({n1}-{n2}))})=\\var{s1}\\)

\n

\\(\\var{n1}a+(\\simplify{({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))})=\\var{s1}\\)

\n

\\(\\var{n1}a=\\var{s1}-(\\simplify{({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))})\\)

\n

\\(\\var{n1}a=\\simplify{{s1}-({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))}\\)

\n

\\(a=\\simplify{({s1}-({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2})))/{n1}}\\)

\n

\\(a=\\var{a}\\)

\n

  

", "type": "question"}, {"name": "Geometric progression: The nth term of a series", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "variables": {"a": {"name": "a", "description": "", "templateType": "randrange", "group": "Ungrouped variables", "definition": "random(1..12#1)"}, "n": {"name": "n", "description": "", "templateType": "randrange", "group": "Ungrouped variables", "definition": "random(4..19#1)"}, "r": {"name": "r", "description": "", "templateType": "randrange", "group": "Ungrouped variables", "definition": "random(0.2..3#0.2)"}}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "statement": "

The first three terms of a series are given by:  

\n

\\(\\var{a} + \\simplify{{a}*{r}} + \\simplify{{a}*{r}^2}\\,+ \\, ...........\\)

", "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Find the nth term of a Geometric progression

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Calculate the \\(\\var{n}th\\) term of the series.

\n

\\(T_\\var{n}=\\) [[0]]

", "scripts": {}}], "advice": "

If the ratio between successive pairs of terms is a constant then the series under examination is a geometric progression.

\n

Ths first term is \\(a\\) and the common ratio is \\(r\\).

\n

The formula for the nth term of the series is given by:    \\(T_n=ar^{n-1}\\)

\n

In this example \\(a=\\var{a}\\),   \\(r = \\frac{\\simplify{{a}*{r}}}{\\var{a}}=\\var{r}\\)  and  \\(n = \\var{n}\\)

\n

\\(T_\\var{n}=\\var{a}*\\var{r}^{\\simplify{{n}-1}}\\)

\n

\\(T_\\var{n}=\\var{a}*\\simplify{{r}^{{n}-1}}\\)

\n

\\(T_\\var{n}=\\simplify{{a}*{r}^{{n}-1}}\\)

", "variablesTest": {"maxRuns": 100, "condition": ""}, "functions": {}, "rulesets": {}, "tags": [], "type": "question"}, {"name": "Geometric progression: The sum of the first n terms of a geometric progression", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Find the sum of the first n terms of a Geometric progression

"}, "advice": "

If the ratio between successive pairs of terms is a constant then the series under examination is a geometric progression.

\n

Ths first term is \\(a\\) and the common ratio is \\(r\\).

\n

The formula for the sum of the first \\(n\\) terms of the series is given by:    \\(S_n=\\frac{a(1-r^{n})}{1-r}\\)

\n

In this example \\(a=\\var{a}\\),   \\(r = \\frac{\\simplify{{a}*{r}}}{\\var{a}}=\\var{r}\\)  and  \\(n = \\var{n}\\)

\n

\\(S_\\var{n}=\\frac{\\var{a}(1-(\\var{r})^{\\var{n}})}{1-\\var{r}}\\)

\n

\\(S_\\var{n}=\\frac{\\var{a}*(\\simplify{1-{r}^{n}})}{\\simplify{1-{r}}}\\)

\n

\\(S_\\var{n}=\\frac{\\simplify{{a}*(1-{r}^{n})}}{\\simplify{1-{r}}}\\)

\n

\\(S_\\var{n}=\\var{s}\\)

", "variable_groups": [], "parts": [{"showFeedbackIcon": true, "marks": 0, "scripts": {}, "variableReplacements": [], "prompt": "

Calculate the sum of the first \\(\\var{n}\\) terms of the series.

\n

\\(S_\\var{n}=\\) [[0]]

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The first three terms of a series are given by:  

\n

\\(\\var{a} + \\simplify{{a}*{r}} + \\simplify{{a}*{r}^2}\\,+ \\, ...........\\)

", "preamble": {"js": "", "css": ""}, "rulesets": {}, "variables": {"n": {"definition": "random(4..19#1)", "templateType": "randrange", "name": "n", "group": "Ungrouped variables", "description": ""}, "a": {"definition": "random(1..12#1)", "templateType": "randrange", "name": "a", "group": "Ungrouped variables", "description": ""}, "s": {"definition": "({a}*(1-{r}^{n}))/(1-{r})", "templateType": "anything", "name": "s", "group": "Ungrouped variables", "description": ""}, "r": {"definition": "random(0.2..3#0.2)", "templateType": "randrange", "name": "r", "group": "Ungrouped variables", "description": ""}}, "ungrouped_variables": ["a", "r", "n", "s"], "functions": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "tags": [], "type": "question"}, {"name": "Solving for a geometric series", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Solving for a geometric series

"}, "advice": "

\\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s_1}\\)

\n

\\(S_{\\simplify{2*{n}}}=\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}=\\var{s_2}\\)

\n

If we divide one by the other we get:

\n

\\(\\frac{S_{\\simplify{2*{n}}}}{S_{\\var{n}}}=\\frac{\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}}{\\frac{a(1-r^{\\var{n}})}{1-r}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

\n

\\(\\frac{S_{\\simplify{2*{n}}}}{S_{\\var{n}}}=\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}*\\frac{1-r}{a(1-r^{\\var{n}})}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

\n

\\(\\frac{1-r^{\\simplify{2*{n}}}}{1-r^{\\var{n}}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

\n

\\(\\frac{(1-r^\\var{n})(1+r^{\\var{n}})}{1-r^{\\var{n}}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

\n

\\(1+r^{\\var{n}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

\n

\\(r^{\\var{n}}=\\frac{\\var{s_2}}{\\var{s_1}}-1\\)

\n

\\(r^{\\var{n}}=\\simplify{{s_2}/{s_1}-1}\\)

\n

\\(r=\\simplify{(({s_2})/{s_1}-1)^{1/{n}}}\\)

\n

\\(r=\\simplify{(({s_2}-{s_1})/{s_1})^{1/{n}}}=\\var{r}\\)

\n

Recall \\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s_1}\\)

\n

\\(a=\\frac{\\var{s_1}*(1-{r})}{1-r^{\\var{n}}}\\)

\n

Inserting the value for \\(r\\) in this equation gives

\n

\\(a=\\frac{\\var{s_1}*(\\simplify{(1-{r})})}{\\simplify{{1-r^{{n}}}}}\\)

\n

\\(a=\\var{a}\\)

\n

", "tags": [], "functions": {}, "statement": "

The sum of the first \\(\\var{n}\\) terms of a geometric series is \\(\\var{s_1}\\) and the sum of the first \\(\\simplify{2*{n}}\\) terms is \\(\\var{s_2}\\).

", "preamble": {"js": "", "css": ""}, "rulesets": {}, "variables": {"n": {"definition": "random(2..6#1)", "templateType": "randrange", "name": "n", "group": "Ungrouped variables", "description": ""}, "a": {"definition": "{s_1}*(1-{r})/(1-{r}^{n})", "templateType": "anything", "name": "a", "group": "Ungrouped variables", "description": ""}, "s_2": {"definition": "random(36..50#1)", "templateType": "randrange", "name": "s_2", "group": "Ungrouped variables", "description": ""}, "s_1": {"definition": "random(12..36#1)", "templateType": "randrange", "name": "s_1", "group": "Ungrouped variables", "description": ""}, "r": {"definition": "({s_2}/{s_1}-1)^(1/{n})", "templateType": "anything", "name": "r", "group": "Ungrouped variables", "description": ""}}, "ungrouped_variables": ["n", "s_1", "s_2", "r", "a"], "variablesTest": {"condition": "", "maxRuns": 100}, "parts": [{"scripts": {}, "showFeedbackIcon": true, "showCorrectAnswer": true, "variableReplacements": [], "prompt": "

Determine the value of the common ratio.    \\(r\\) = [[0]]

\n

Calculate the value of the first term.    \\(a\\) = [[1]]

", "type": "gapfill", "marks": 0, "gaps": [{"minValue": "{r}", "precision": "2", "scripts": {}, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showPrecisionHint": true, "allowFractions": false, "precisionPartialCredit": 0, "marks": 1, "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "precisionType": "dp", "maxValue": "{r}", "correctAnswerFraction": false, "variableReplacements": [], "strictPrecision": false, "mustBeReduced": false, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "mustBeReducedPC": 0}, {"minValue": "{a}", "precision": "2", "scripts": {}, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showPrecisionHint": true, "allowFractions": false, "precisionPartialCredit": 0, "marks": 1, "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "precisionType": "dp", "maxValue": "{a}", "correctAnswerFraction": false, "variableReplacements": [], "strictPrecision": false, "mustBeReduced": false, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "mustBeReducedPC": 0}], "variableReplacementStrategy": "originalfirst"}], "variable_groups": [], "type": "question"}, {"name": "Solving for a geometric series #2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "rulesets": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"js": "", "css": ""}, "tags": [], "metadata": {"description": "

Solving for a geometric series

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "variables": {"r": {"description": "", "name": "r", "definition": "(1-s1/s2)^(1/{n})", "templateType": "anything", "group": "Ungrouped variables"}, "s1": {"description": "", "name": "s1", "definition": "random(12..50#1)", "templateType": "randrange", "group": "Ungrouped variables"}, "s2": {"description": "", "name": "s2", "definition": "2*{s1}-{n}", "templateType": "anything", "group": "Ungrouped variables"}, "a": {"description": "", "name": "a", "definition": "{s2}*(1-{r})", "templateType": "anything", "group": "Ungrouped variables"}, "n": {"description": "", "name": "n", "definition": "random(5..12#1)", "templateType": "randrange", "group": "Ungrouped variables"}}, "parts": [{"marks": 0, "type": "gapfill", "showCorrectAnswer": true, "scripts": {}, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "variableReplacements": [], "prompt": "

Determine the value of the common ratio.    \\(r\\) = [[0]]

\n

Calculate the value of the first term.    \\(a\\) = [[1]]

", "gaps": [{"marks": 1, "showPrecisionHint": true, "mustBeReduced": false, "maxValue": "{r}", "notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "correctAnswerFraction": false, "variableReplacements": [], "type": "numberentry", "scripts": {}, "allowFractions": false, "precisionPartialCredit": 0, "minValue": "{r}", "correctAnswerStyle": "plain", "showCorrectAnswer": true, "precisionType": "dp", "mustBeReducedPC": 0, "precision": "2", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "precisionMessage": "You have not given your answer to the correct precision."}, {"marks": 1, "showPrecisionHint": true, "mustBeReduced": false, "maxValue": "{a}", "notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "correctAnswerFraction": false, "variableReplacements": [], "type": "numberentry", "scripts": {}, "allowFractions": false, "precisionPartialCredit": 0, "minValue": "{a}", "correctAnswerStyle": "plain", "showCorrectAnswer": true, "precisionType": "dp", "mustBeReducedPC": 0, "precision": "2", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "precisionMessage": "You have not given your answer to the correct precision."}]}], "statement": "

The sum of the first \\(\\var{n}\\) terms of a geometric series is \\(\\var{s1}\\) and the sum to infinity of the series is \\(\\var{s2}\\).

", "variable_groups": [], "functions": {}, "advice": "

\\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s1}\\)

\n

\\(S_{\\infty}=\\frac{a}{1-r}=\\var{s2}\\)

\n

If we divide one by the other we get:

\n

\\(\\frac{S_{\\simplify{n}}}{S_{\\infty}}=\\frac{\\frac{a(1-r^{\\simplify{n}})}{1-r}}{\\frac{a}{1-r}}=\\frac{\\var{s1}}{\\var{s2}}\\)

\n

\\(\\frac{S_{\\simplify{n}}}{S_{\\infty}}=\\frac{a(1-r^{\\simplify{n}})}{1-r}*\\frac{1-r}{a}=\\frac{\\var{s1}}{\\var{s2}}\\)

\n

\\(1-r^{{n}}=\\frac{\\var{s1}}{\\var{s2}}\\)

\n

\\(1-\\frac{\\var{s1}}{\\var{s2}}=r^{{n}}\\)

\n

In this example \\(n=\\var{n}\\)

\n

\\(r^{\\var{n}}=\\simplify{({s2}-{s1})/{s2}}\\)

\n

\\(r=(\\simplify{({s2}-{s1})/{s2}})^{1/\\var{n}}\\)

\n

\\(r=\\simplify{(({s2}-{s1})/{s1})^{1/{n}}}=\\var{r}\\)

\n

Recall \\(S_{\\infty}=\\frac{a}{1-r}=\\var{s2}\\)

\n

\\(a=\\var{s2}*(1-{r})\\)

\n

Inserting the value for \\(r\\) in this equation gives

\n

\\(a=\\var{s2}*\\simplify{(1-{r})}\\)

\n

\\(a=\\var{a}\\)

\n

", "ungrouped_variables": ["n", "s1", "s2", "r", "a"], "type": "question"}, {"name": "Solving for a geometric series #3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Solving for a geometric series

"}, "advice": "

The second term of a geometric series is given by the formula \\(T_2=ar\\) and the sum to infinity of a geometric series is \\(S_\\infty=\\frac{a}{1-r}\\)

\n

\\(T_2=ar=\\var{t2}\\)

\n

\\(a=\\frac{\\var{t2}}{r}\\)

\n

We can substitute this in for \\(a\\) in the second equation

\n

\\(S_\\infty=\\frac{a}{1-r}=\\var{s}\\)

\n

\\(\\frac{\\frac{\\var{t2}}{r}}{1-r}=\\var{s}\\)

\n

\\(\\frac{\\var{t2}}{r}=\\var{s}(1-{r})\\)

\n

\\(\\frac{\\var{t2}}{r}=\\var{s}-\\var{s}{r}\\)

\n

\\(\\var{t2}=\\var{s}r-\\var{s}r^2\\)

\n

\\(\\var{s}r^2-\\var{s}r+\\var{t2}=0\\) 

\n

This is a quadratic equation which we can solve by formula.

\n

\\(r=\\frac{\\var{s}\\pm \\sqrt{(-\\var{s})^2-4*(\\var{s})*(\\var{t2})}}{2*(\\var{s})}\\)

\n

\\(r=\\frac{\\var{s}+\\sqrt{\\simplify{{s}^2-4*{s}*{t2}}}}{\\simplify{2*{s}}}\\)   or    \\(r=\\frac{\\var{s}-\\sqrt{\\simplify{{s}^2-4*{s}*{t2}}}}{\\simplify{2*{s}}}\\)

\n

\\(r=\\frac{\\var{s}+\\simplify{({s}^2-4*{s}*{t2})^0.5}}{\\simplify{2*{s}}}\\)    or    \\(r=\\frac{\\var{s}-\\simplify{({s}^2-4*{s}*{t2})^0.5}}{\\simplify{2*{s}}}\\)

\n

\\(r=\\) {({s}+({s}^2-4*{s}*{t2})^0.5)/(2*{s})}   or    \\(r=\\) {({s}-({s}^2-4*{s}*{t2})^0.5)/(2*{s})}

\n

\\(a=\\frac{\\var{t2}}{r}\\)

\n

\\(a=\\) {(2*{s}*{t2})/({s}+({s}^2-4*{s}*{t2})^0.5)}    or    \\(a=\\) {(2*{s}*{t2})/({s}-({s}^2-4*{s}*{t2})^0.5)}

\n

\n

\n

\n

\n

", "variable_groups": [], "parts": [{"scripts": {}, "showFeedbackIcon": true, "prompt": "

Calculate the value of the larger common ratio.   \\(r\\) = [[0]]

\n

Determine the value of the first term of the series corresponding to this common ratio.  \\(a\\) = [[1]]

\n

Calculate the value of the smaller common ratio.   \\(r\\) = [[2]]

\n

Determine the value of the first term of the series corresponding to this common ratio.  \\(a\\) = [[3]]

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The second term in a geometric series is \\(\\var{t2}\\) and the sum to infinity of the series is \\(\\var{s}\\).

\n

There are two possible series that possess these attributes.

", "functions": {}, "preamble": {"js": "", "css": ""}, "rulesets": {}, "variables": {"a_2": {"definition": "{t2}/{r_2}", "templateType": "anything", "name": "a_2", "group": "Ungrouped variables", "description": ""}, "r_2": {"definition": "({s}-sqrt({s}^2-4*{s}*{t2}))/(2*{s})", "templateType": "anything", "name": "r_2", "group": "Ungrouped variables", "description": ""}, "t2": {"definition": "random(1..9#1)", "templateType": "randrange", "name": "t2", "group": "Ungrouped variables", "description": ""}, "a_1": {"definition": "{t2}/{r_1}", "templateType": "anything", "name": "a_1", "group": "Ungrouped variables", "description": ""}, "s": {"definition": "random(36..50#1)", "templateType": "randrange", "name": "s", "group": "Ungrouped variables", "description": ""}, "r_1": {"definition": "({s}+sqrt({s}^2-4*{s}*{t2}))/(2*{s})", "templateType": "anything", "name": "r_1", "group": "Ungrouped variables", "description": ""}}, "ungrouped_variables": ["t2", "s", "r_1", "r_2", "a_1", "a_2"], "variablesTest": {"condition": "", "maxRuns": 100}, "tags": [], "type": "question"}, {"name": "First 3 terms of Binomial series for Natural exponent", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "variables": {"k": {"description": "", "templateType": "anything", "name": "k", "group": "Ungrouped variables", "definition": "random(2..{n}-1)"}, "a": {"description": "", "templateType": "randrange", "name": "a", "group": "Ungrouped variables", "definition": "random(2..10#1)"}, "n": {"description": "", "templateType": "randrange", "name": "n", "group": "Ungrouped variables", "definition": "random(4..9#1)"}, "c": {"description": "", "templateType": "anything", "name": "c", "group": "Ungrouped variables", "definition": "comb({n},{k})*{a}^({n}-{k})*{b}^{k}"}, "b": {"description": "", "templateType": "randrange", "name": "b", "group": "Ungrouped variables", "definition": "random(2..12#1)"}}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "statement": "

Given the expression \\((\\var{a}+\\var{b}x)^{\\var{n}}\\)

", "metadata": {"description": "

Find the first 3 terms of Binomial series having a Natural exponent

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "advice": "

The binomial series expansion for an expression of the form \\((a+bx)^n\\) where \\(n\\) is a Natural number is given by:

\n

\\((a+bx)^n=\\tbinom{n}{0}(a)^n(bx)^{0}+\\tbinom{n}{1}(a)^{n-1}(bx)^{1}+\\tbinom{n}{2}(a)^{n-2}(bx)^{2}+...\\tbinom{n}{k}(a)^{n-k}(bx)^{k}+...\\tbinom{n}{n}(a)^{0}(bx)^{n}\\)

\n

In this example  \\(n=\\var{n}\\),  \\(k=\\var{k}\\),  \\(a=\\var{a}\\)  and  \\(b=\\var{b}\\).

\n

So the first three terms of the binomial series expansion are:

\n

 \\(\\var{a}^{\\var{n}}+\\tbinom{\\var{n}}{\\var{1}}*\\var{a}^{\\var{n}-1}*\\var{b}^{1}+\\tbinom{\\var{n}}{2}*\\var{a}^{\\var{n}-2}*\\var{b}^{2}\\)

\n

\\(=\\simplify{{a}^{n}}+\\simplify{{n}*{a}^({n}-1)*{b}}x+\\simplify{{n}*{n-1}*{a}^{{n}-2}*{b}^2/2x^2}\\)

\n

", "parts": [{"showFeedbackIcon": true, "marks": "3", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "checkvariablenames": false, "prompt": "

Write out the first three terms of the binomial expansion.  

", "scripts": {}, "answer": "{a}^{n}+{n}*{a}^({n}-1)*{b}x+{n}*{n-1}*{a}^({n}-2)*{b}^2/2x^2", "showCorrectAnswer": true, "vsetrange": [0, 1], "checkingtype": "absdiff", "expectedvariablenames": [], "type": "jme", "vsetrangepoints": 5, "checkingaccuracy": 0.001, "showpreview": true}], "ungrouped_variables": ["a", "b", "n", "c", "k"], "variablesTest": {"maxRuns": 100, "condition": ""}, "tags": [], "rulesets": {}, "functions": {}, "type": "question"}, {"name": "Binomial series for Natural exponent", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "variablesTest": {"condition": "", "maxRuns": 100}, "preamble": {"js": "", "css": ""}, "tags": [], "statement": "

Given the expression \\((\\var{a}+\\var{b}x)^{\\var{n}}\\)

", "parts": [{"prompt": "

By using the binomial series expansion, calculate the coefficient of \\(x^{\\var{k}}\\)  [[0]]

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The binomial series expansion for an expression of the form \\((a+bx)^n\\) where \\(n\\) is a Natural number is given by:

\n

\\((a+bx)^n=\\tbinom{n}{0}(a)^n(bx)^{0}+\\tbinom{n}{1}(a)^{n-1}(bx)^{1}+\\tbinom{n}{2}(a)^{n-2}(bx)^{2}+...\\tbinom{n}{k}(a)^{n-k}(bx)^{k}+...\\tbinom{n}{n}(a)^{0}(bx)^{n}\\)

\n

In this example  \\(n=\\var{n}\\),  \\(k=\\var{k}\\),  \\(a=\\var{a}\\)  and  \\(b=\\var{b}\\).

\n

So the coefficient of \\(x^{\\var{k}}\\) is given by \\(\\tbinom{\\var{n}}{\\var{k}}*\\var{a}^{\\var{n}-\\var{k}}*\\var{b}^{\\var{k}}=\\var{c}\\).

\n

", "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Binomial series for Natural exponent

"}, "ungrouped_variables": ["a", "b", "n", "c", "k"], "functions": {}, "type": "question"}, {"name": " Binomial series for non-Natural exponent", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "ungrouped_variables": ["b", "n", "T1", "T2", "T3"], "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Find the first three terms in the binomial series expansion having a non-Natural exponent

"}, "tags": [], "functions": {}, "parts": [{"showCorrectAnswer": true, "showFeedbackIcon": true, "checkingtype": "absdiff", "checkingaccuracy": 0.001, "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "checkvariablenames": false, "showpreview": true, "vsetrange": [0, 1], "type": "jme", "answer": "{T1}+{T2}x+{T3}x^2", "prompt": "

Input the first three tems in the binomial series expansion. 

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The binomial series expansion for an expression of the form \\((1+bx)^{n}\\) where \\(n\\) is a non-natural number, is given by:

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\\((1+bx)^{n}=1+\\tbinom{n}{1}(bx)^{1}+\\tbinom{{n}}{2}(bx)^{2}+\\tbinom{{n}}{3}(bx)^{3}.......\\)

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In this example  \\(power=\\frac{1}{\\var{n}}\\),   and  \\(b=\\var{b}\\).

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So the first three terms are

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\\((1+\\var{b}x)^{\\frac{1}{\\var{n}}}=1+\\tbinom{\\frac{1}{\\var{n}}}{1}(\\var{b}x)^{1}+\\tbinom{\\frac{1}{\\var{n}}}{2}(\\var{b}x)^{2}\\)

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\\((1+\\var{b}x)^{\\frac{1}{\\var{n}}}=\\var{T1}+\\var{T2}x+\\var{T3}x^2\\)

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Given the expression \\((1+\\var{b}x)^{\\frac{1}{\\var{n}}}\\)

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This quiz poses questions on arithmetic progressions, geometric progressions and binomial series expansions.

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