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Arithmetico-geometric sequence
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<h2 id="elements">Elements</h2> <p>The elements of an arithmetico-geometric sequence ( A n G n ) n ≥ 1 {\displaystyle (A_{n}G_{n})_{n\geq 1}} are the products of the elements of an <a href="/facts/Arithmetic_progression/c4ew9Xoq">arithmetic progression</a> ( A n ) n ≥ 1 {\displaystyle (A_{n})_{n\geq 1}} (in blue) with initial value a {\displaystyle a} and common difference d {\displaystyle d} , A n = a + ( n − 1 ) d , {\displaystyle A_{n}=a+(n-1)d,} with the corresponding elements of a <a href="/facts/Geometric_progression/SVrkRHDX">geometric progression</a> ( G n ) n ≥ 1 {\displaystyle (G_{n})_{n\geq 1}} (in green) with initial value b {\displaystyle b} and common ratio r {\displaystyle r} , G n = b r n − 1 , {\displaystyle G_{n}=br^{n-1},} so that<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a> </p> A 1 G 1 = a b A 2 G 2 = ( a + d ) b r A 3 G 3 = ( a + 2 d ) b r 2 ⋮ A n G n = ( a + ( n − 1 ) d ) b r n − 1 . {\displaystyle {\begin{aligned}A_{1}G_{1}&=\color {blue}a\color {green}b\\A_{2}G_{2}&=\color {blue}(a+d)\color {green}br\\A_{3}G_{3}&=\color {blue}(a+2d)\color {green}br^{2}\\&\ \,\vdots \\A_{n}G_{n}&=\color {blue}{\bigl (}a+(n-1)d{\bigr )}\color {green}br^{n-1}\color {black}.\end{aligned}}} <p>These four parameters are somewhat redundant and can be reduced to three: a b , {\displaystyle ab,} b d , {\displaystyle bd,} and r . {\displaystyle r.} </p> <h3>Example</h3> <p>The sequence </p> 0 1 , 1 2 , 2 4 , 3 8 , 4 16 , 5 32 , ⋯ {\displaystyle {\frac {\color {blue}{0}}{\color {green}{1}}},\ {\frac {\color {blue}{1}}{\color {green}{2}}},\ {\frac {\color {blue}{2}}{\color {green}{4}}},\ {\frac {\color {blue}{3}}{\color {green}{8}}},\ {\frac {\color {blue}{4}}{\color {green}{16}}},\ {\frac {\color {blue}{5}}{\color {green}{32}}},\cdots } <p>is the arithmetico-geometric sequence with parameters d = b = 1 {\displaystyle d=b=1} , a = 0 {\displaystyle a=0} , and r = 1 2 {\displaystyle r={\tfrac {1}{2}}} . </p> <h2 id="series">Series</h2> <h3>Partial sums</h3> <p>The sum of the first <i>n</i> terms of an arithmetico-geometric series has the form </p> S n = ∑ k = 1 n A k G k = ∑ k = 1 n ( a + ( k − 1 ) d ) b r k − 1 = b ∑ k = 0 n − 1 ( a + k d ) r k = a b + ( a + d ) b r + ( a + 2 d ) b r 2 + ⋯ + ( a + ( n − 1 ) d ) b r n − 1 {\displaystyle {\begin{aligned}S_{n}&=\sum _{k=1}^{n}A_{k}G_{k}\\[5pt]&=\sum _{k=1}^{n}{\bigl (}a+(k-1)d{\bigr )}br^{k-1}\\[5pt]&=b\sum _{k=0}^{n-1}\left(a+kd\right)r^{k}\\[5pt]&=ab+(a+d)br+(a+2d)br^{2}+\cdots +{\bigl (}a+(n-1)d{\bigr )}br^{n-1}\end{aligned}}} <p>where A i {\textstyle A_{i}} and G i {\textstyle G_{i}} are the ith elements of the arithmetic and the geometric sequence, respectively. </p><p>This partial sum has the <a href="/facts/Closed-form_expression/y0xCXlSk">closed-form expression</a> </p> S n = a b − ( a + n d ) b r n 1 − r + d b r ( 1 − r n ) ( 1 − r ) 2 = A 1 G 1 − A n + 1 G n + 1 1 − r + d r ( 1 − r ) 2 ( G 1 − G n + 1 ) . {\displaystyle {\begin{aligned}S_{n}&={\frac {ab-(a+nd)\,br^{n}}{1-r}}+{\frac {dbr\,(1-r^{n})}{(1-r)^{2}}}\\&={\frac {A_{1}G_{1}-A_{n+1}G_{n+1}}{1-r}}+{\frac {dr}{(1-r)^{2}}}\,(G_{1}-G_{n+1}).\end{aligned}}} <h4>Derivation</h4> <p>Multiplying<a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a> </p> S n = a b + ( a + d ) b r + ( a + 2 d ) b r 2 + ⋯ + ( a + ( n − 1 ) d ) b r n − 1 {\displaystyle S_{n}=ab+(a+d)br+(a+2d)br^{2}+\cdots +{\bigl (}a+(n-1)d{\bigr )}br^{n-1}} <p>by <i>r</i> gives </p> r S n = a b r + ( a + d ) b r 2 + ( a + 2 d ) b r 3 + ⋯ + ( a + ( n − 1 ) d ) b r n . {\displaystyle rS_{n}=abr+(a+d)br^{2}+(a+2d)br^{3}+\cdots +{\bigl (}a+(n-1)d{\bigr )}br^{n}.} <p>Subtracting <i>rSn</i> from <i>Sn</i>, dividing both sides by b {\displaystyle b} , and using the technique of <a href="/facts/Telescoping_series/9QHSco21">telescoping series</a> (second equality) and the formula for the sum of a finite <a href="/facts/Geometric_series/Ne5TBH57">geometric series</a> (fifth equality) gives </p> ( 1 − r ) S n b = ( a + ( a + d ) r + ( a + 2 d ) r 2 + ⋯ + ( a + ( n − 1 ) d ) r n − 1 ) − ( a r + ( a + d ) r 2 + ( a + 2 d ) r 3 + ⋯ + ( a + ( n − 1 ) d ) r n ) = a + d ( r + r 2 + ⋯ + r n − 1 ) − ( a + ( n − 1 ) d ) r n = a + d ( r + r 2 + ⋯ + r n − 1 + r n ) − ( a + n d ) r n = a + d r ( 1 + r + r 2 + ⋯ + r n − 1 ) − ( a + n d ) r n = a + d r ( 1 − r n ) 1 − r − ( a + n d ) r n , S n = b 1 − r ( a − ( a + n d ) r n + d r ( 1 − r n ) 1 − r ) = a b − ( a + n d ) b r n 1 − r + d r ( b − b r n ) ( 1 − r ) 2 = A 1 G 1 − A n + 1 G n + 1 1 − r + d r ( G 1 − G n + 1 ) ( 1 − r ) 2 {\displaystyle {\begin{aligned}{\frac {(1-r)S_{n}}{b}}&=\left(a+(a+d)r+(a+2d)r^{2}+\cdots +{\bigl (}a+(n-1)d{\bigr )}r^{n-1}\right)-{\Bigl (}ar+(a+d)r^{2}+(a+2d)r^{3}+\cdots +{\bigl (}a+(n-1)d{\bigr )}r^{n}{\Bigr )}\\[5pt]&=a+d\left(r+r^{2}+\cdots +r^{n-1}\right)-{\bigl (}a+(n-1)d{\bigr )}r^{n}\\[5pt]&=a+d\left(r+r^{2}+\cdots +r^{n-1}+r^{n}\right)-\left(a+nd\right)r^{n}\\[5pt]&=a+dr\left(1+r+r^{2}+\cdots +r^{n-1}\right)-\left(a+nd\right)r^{n}\\[5pt]&=a+{\frac {dr(1-r^{n})}{1-r}}-(a+nd)r^{n},\\[8pt]S_{n}&={\frac {b}{1-r}}\left(a-(a+nd)r^{n}+{\frac {dr(1-r^{n})}{1-r}}\right)\\[5pt]&={\frac {ab-(a+nd)br^{n}}{1-r}}+{\frac {dr(b-br^{n})}{(1-r)^{2}}}\\[5pt]&={\frac {A_{1}G_{1}-A_{n+1}G_{n+1}}{1-r}}+{\frac {dr(G_{1}-G_{n+1})}{(1-r)^{2}}}\end{aligned}}} <p>as claimed. </p> <h3>Infinite series</h3> <p>If −1 < <i>r</i> < 1, then the sum <i>S</i> of the arithmetico-geometric <a href="/facts/Series_(mathematics)/yDnlsgIe">series</a>, that is to say, the <a href="/facts/Limit_of_a_sequence/Ktge2RwL">limit</a> of the partial sums of the elements of the sequence, is given by<a class="footnote-ref" id="fnref:6" href="#fn:6"><sup>6</sup></a> </p> S = ∑ k = 1 ∞ t k = lim n → ∞ S n = a b 1 − r + d b r ( 1 − r ) 2 = A 1 G 1 1 − r + d r G 1 ( 1 − r ) 2 . {\displaystyle {\begin{aligned}S&=\sum _{k=1}^{\infty }t_{k}=\lim _{n\to \infty }S_{n}\\[5pt]&={\frac {ab}{1-r}}+{\frac {dbr}{(1-r)^{2}}}\\[5pt]&={\frac {A_{1}G_{1}}{1-r}}+{\frac {drG_{1}}{(1-r)^{2}}}.\end{aligned}}} <p>If <i>r</i> is outside of the above range, <i>b</i> is not zero, and <i>a</i> and <i>d</i> are not both zero, the limit does not exist and the series is <a href="/facts/Divergent_series/c8Vjjjdx">divergent</a>. </p> <h3>Example</h3> <p>The sum </p> S = 0 1 + 1 2 + 2 4 + 3 8 + 4 16 + 5 32 + ⋯ {\displaystyle S={\frac {\color {blue}{0}}{\color {green}{1}}}+{\frac {\color {blue}{1}}{\color {green}{2}}}+{\frac {\color {blue}{2}}{\color {green}{4}}}+{\frac {\color {blue}{3}}{\color {green}{8}}}+{\frac {\color {blue}{4}}{\color {green}{16}}}+{\frac {\color {blue}{5}}{\color {green}{32}}}+\cdots } , <p>is the sum of an arithmetico-geometric series defined by d = b = 1 {\displaystyle d=b=1} , a = 0 {\displaystyle a=0} , and r = 1 2 {\displaystyle r={\tfrac {1}{2}}} , and it converges to S = 2 {\displaystyle S=2} . This sequence corresponds to the expected number of <a href="/facts/Coin_flipping/uAJd6pAJ">coin tosses</a> required to obtain "tails". The probability T k {\displaystyle T_{k}} of obtaining tails for the first time at the <i>k</i>th toss is as follows: </p> T 1 = 1 2 , T 2 = 1 4 , … , T k = 1 2 k {\displaystyle T_{1}={\frac {1}{2}},\ T_{2}={\frac {1}{4}},\dots ,T_{k}={\frac {1}{2^{k}}}} . <p>Therefore, the expected number of tosses to reach the first "tails" is given by </p> ∑ k = 1 ∞ k T k = ∑ k = 1 ∞ k 2 k = 2. {\displaystyle \sum _{k=1}^{\infty }kT_{k}=\sum _{k=1}^{\infty }{\frac {\color {blue}k}{\color {green}2^{k}}}=2.} <p>Similarly, the sum </p> S = 0 ⋅ 1 6 5 6 + 1 ⋅ 1 6 1 + 2 ⋅ 1 6 6 5 + 3 ⋅ 1 6 ( 6 5 ) 2 + 4 ⋅ 1 6 ( 6 5 ) 3 + 5 ⋅ 1 6 ( 6 5 ) 4 + ⋯ {\displaystyle S={\frac {\color {blue}{0}\cdot \color {green}{\frac {1}{6}}}{\color {green}{\frac {5}{6}}}}+{\frac {\color {blue}{1}\cdot \color {green}{\frac {1}{6}}}{\color {green}{1}}}+{\frac {\color {blue}{2}\cdot \color {green}{\frac {1}{6}}}{\color {green}{\frac {6}{5}}}}+{\frac {\color {blue}{3}\cdot \color {green}{\frac {1}{6}}}{\color {green}{\left({\frac {6}{5}}\right)^{2}}}}+{\frac {\color {blue}{4}\cdot \color {green}{\frac {1}{6}}}{\color {green}{\left({\frac {6}{5}}\right)^{3}}}}+{\frac {\color {blue}{5}\cdot \color {green}{\frac {1}{6}}}{\color {green}{\left({\frac {6}{5}}\right)^{4}}}}+\cdots } <p>is the sum of an arithmetico-geometric series defined by d = 1 {\displaystyle d=1} , a = 0 {\displaystyle a=0} , b = 1 / 6 5 / 6 = 1 5 {\displaystyle b={\tfrac {1/6}{5/6}}={\tfrac {1}{5}}} , and r = 5 6 {\displaystyle r={\tfrac {5}{6}}} , and it converges to 6. This sequence corresponds to the expected number of <a href="/facts/Six-sided_dice/qBnylzwJ">six-sided dice</a> rolls required to obtain a specific value on a die roll, for instance "5". In general, these series with d = 1 {\displaystyle d=1} , a = 0 {\displaystyle a=0} , b = p 1 − p {\displaystyle b={\tfrac {p}{1-p}}} , and r = 1 − p {\displaystyle r=1-p} give the expectations of "the number of trials until first success" in <a href="/facts/Bernoulli_process/AF9E3jpu">Bernoulli processes</a> with "success probability" p {\displaystyle p} . The probabilities of each outcome follow a <a href="/facts/Geometric_distribution/c8wjfaoV">geometric distribution</a> and provide the geometric sequence factors in the terms of the series, while the number of trials per outcome provides the arithmetic sequence factors in the terms. </p> <h2 id="further-reading">Further reading</h2> <ul><li>D. Khattar. <a href="https://books.google.com/books?id=5OffscY1FGYC&pg=SA10-PA8"><i>The Pearson Guide to Mathematics for the IIT-JEE, 2/e</i></a> (New ed.). Pearson Education India. p. 10.8. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 81-317-2876-5.</li> <li>P. Gupta. <a href="https://books.google.com/books?id=rQtlgvS598MC&pg=PA380"><i>Comprehensive Mathematics XI</i></a>. Laxmi Publications. p. 380. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 81-7008-597-7.</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>"Arithmetic-Geometric Progression | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2021-04-21. <a href="https://brilliant.org/wiki/arithmetic-geometric-progression/" target="_blank">https://brilliant.org/wiki/arithmetic-geometric-progression/</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Swain, Stuart G. (2018). "Proof Without Words: Gabriel's Staircase". Mathematics Magazine. 67 (3): 209. doi:10.1080/0025570X.1994.11996214. ISSN 0025-570X. <a href="/wiki/Doi_(identifier)" target="_blank">/wiki/Doi_(identifier)</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>Edgar, Tom (2018). "Staircase Series". Mathematics Magazine. 91 (2): 92–95. doi:10.1080/0025570X.2017.1415584. ISSN 0025-570X. S2CID 218542483. <a href="/wiki/Doi_(identifier)" target="_blank">/wiki/Doi_(identifier)</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>K. F. Riley; M. P. Hobson; S. J. Bence (2010). Mathematical methods for physics and engineering (3rd ed.). Cambridge University Press. p. 118. ISBN 978-0-521-86153-3. <a href="978-0-521-86153-3" target="_blank">978-0-521-86153-3</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>K. F. Riley; M. P. Hobson; S. J. Bence (2010). Mathematical methods for physics and engineering (3rd ed.). Cambridge University Press. p. 118. ISBN 978-0-521-86153-3. <a href="978-0-521-86153-3" target="_blank">978-0-521-86153-3</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> <li id="fn:6"><p>K. F. Riley; M. P. Hobson; S. J. Bence (2010). Mathematical methods for physics and engineering (3rd ed.). Cambridge University Press. p. 118. ISBN 978-0-521-86153-3. <a href="978-0-521-86153-3" target="_blank">978-0-521-86153-3</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li> </ol>