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Q-An-AP-has-even-number-of-terms-the-sum-of-odd-terms-is-24-and-of-even-terms-is-30-the-last-term-exceeds-the-first-term-by-10-1-2-find-AP-and-number-of-terms-

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Question Number 7395 by 2402@gmail.com last updated on 26/Aug/16
Q. An AP has even number of terms the sum of odd terms is 24 and of even terms is 30. the last term exceeds the first term by 10(1/2) . find AP and number of terms.
$${Q}.\:{An}\:{AP}\:{has}\:{even}\:{number}\:{of}\:{terms} \\ $$$$\:\:\:\:\:\:\:{the}\:{sum}\:{of}\:{odd}\:{terms}\:{is}\:\mathrm{24}\:{and}\: \\ $$$$\:\:\:\:\:\:{of}\:{even}\:{terms}\:{is}\:\mathrm{30}.\:{the}\:{last}\:{term} \\ $$$$\:\:\:\:\:{exceeds}\:{the}\:{first}\:{term}\:{by}\:\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$$$\:\:\:\:\:{find}\:{AP}\:{and}\:{number}\:{of}\:{terms}. \\ $$
Commented by Yozzia last updated on 26/Aug/16
Let n=2k be number of terms. Let u(t)=a+(t−1)d for fixed a and d. All even terms are u(2),u(4),u(6)...u(2t),...u(2k). ∴u(2t)=a+(2t−1)d=h(t) All odd terms are u(1),u(3),u(5),...,u(2t−1),...u(2k−1). ∴u(2t−1)=a+(2t−2)d=a+2(t−1)d=g(t). Now, Σ_(t=1) ^k g(t)=24 and Σ_(t=1) ^k h(t)=30. Σ_(t=1) ^k g(t)=24 ⇔ Σ_(t=1) ^k (a+2d(t−1))=24 ⇔Σ_(t=1) ^k (a−2d)+2dΣ_(t=1) ^k t=(a−2d)k+dk(k+1)=24 or 24=k(a+dk−d) or 24=k(a+dk)−kd....(1) Σ_(t=1) ^k h(t)=30⇔Σ_(t=1) ^k (a+(2t−1)d)=30 ⇔Σ_(t=1) ^k (a−d)+2dΣ_(t=1) ^k t=30⇔k(a−d)+dk(k+1)=30 or k(a+dk)=30......(2) Comparing (1) and (2), (1) becomes 24=30−kd⇒kd=6 or k=(6/d)....(3) The last term is given by u(2k)=a+(2k−1)d and u(2k)−a=((21)/2). ∴ (2k−1)d=((21)/2). But from (3), k=(6/d). ∴ d(((12)/d)−1)=((21)/2) 12−d=((21)/2)⇒d=12−((21)/2)=((24−21)/2)=(3/2). ∴k=(6/(3/2))=4. In (2) ⇒4(a+4×(3/2))=30 4a+24=30 or a=(3/2). The AP is hence the sequence {u(t)}_(t=1) ^8 where u(t)=(3/2)(1+t−1)=((3t)/2), and the number of terms is 2k=2×4=8.
$${Let}\:{n}=\mathrm{2}{k}\:{be}\:{number}\:{of}\:{terms}. \\ $$$${Let}\:{u}\left({t}\right)={a}+\left({t}−\mathrm{1}\right){d}\:{for}\:{fixed}\:{a}\:{and}\:{d}. \\ $$$${All}\:{even}\:{terms}\:{are}\:{u}\left(\mathrm{2}\right),{u}\left(\mathrm{4}\right),{u}\left(\mathrm{6}\right)…{u}\left(\mathrm{2}{t}\right),…{u}\left(\mathrm{2}{k}\right). \\ $$$$\therefore{u}\left(\mathrm{2}{t}\right)={a}+\left(\mathrm{2}{t}−\mathrm{1}\right){d}={h}\left({t}\right) \\ $$$${All}\:{odd}\:{terms}\:{are}\:{u}\left(\mathrm{1}\right),{u}\left(\mathrm{3}\right),{u}\left(\mathrm{5}\right),…,{u}\left(\mathrm{2}{t}−\mathrm{1}\right),…{u}\left(\mathrm{2}{k}−\mathrm{1}\right). \\ $$$$\therefore{u}\left(\mathrm{2}{t}−\mathrm{1}\right)={a}+\left(\mathrm{2}{t}−\mathrm{2}\right){d}={a}+\mathrm{2}\left({t}−\mathrm{1}\right){d}={g}\left({t}\right). \\ $$$${Now},\:\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}{g}\left({t}\right)=\mathrm{24}\:{and}\:\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}{h}\left({t}\right)=\mathrm{30}. \\ $$$$\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}{g}\left({t}\right)=\mathrm{24}\:\Leftrightarrow\:\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}\left({a}+\mathrm{2}{d}\left({t}−\mathrm{1}\right)\right)=\mathrm{24} \\ $$$$\Leftrightarrow\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}\left({a}−\mathrm{2}{d}\right)+\mathrm{2}{d}\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}{t}=\left({a}−\mathrm{2}{d}\right){k}+{dk}\left({k}+\mathrm{1}\right)=\mathrm{24} \\ $$$${or}\:\mathrm{24}={k}\left({a}+{dk}−{d}\right)\:{or}\:\mathrm{24}={k}\left({a}+{dk}\right)−{kd}….\left(\mathrm{1}\right) \\ $$$$\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}{h}\left({t}\right)=\mathrm{30}\Leftrightarrow\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}\left({a}+\left(\mathrm{2}{t}−\mathrm{1}\right){d}\right)=\mathrm{30} \\ $$$$\Leftrightarrow\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}\left({a}−{d}\right)+\mathrm{2}{d}\underset{{t}=\mathrm{1}} {\overset{{k}} {\sum}}{t}=\mathrm{30}\Leftrightarrow{k}\left({a}−{d}\right)+{dk}\left({k}+\mathrm{1}\right)=\mathrm{30} \\ $$$${or}\:{k}\left({a}+{dk}\right)=\mathrm{30}……\left(\mathrm{2}\right) \\ $$$${Comparing}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right),\:\left(\mathrm{1}\right)\:{becomes} \\ $$$$\mathrm{24}=\mathrm{30}−{kd}\Rightarrow{kd}=\mathrm{6}\:{or}\:{k}=\frac{\mathrm{6}}{{d}}….\left(\mathrm{3}\right) \\ $$$${The}\:{last}\:{term}\:{is}\:{given}\:{by}\:{u}\left(\mathrm{2}{k}\right)={a}+\left(\mathrm{2}{k}−\mathrm{1}\right){d} \\ $$$${and}\:{u}\left(\mathrm{2}{k}\right)−{a}=\frac{\mathrm{21}}{\mathrm{2}}.\:\therefore\:\left(\mathrm{2}{k}−\mathrm{1}\right){d}=\frac{\mathrm{21}}{\mathrm{2}}. \\ $$$${But}\:{from}\:\left(\mathrm{3}\right),\:{k}=\frac{\mathrm{6}}{{d}}.\:\therefore\:{d}\left(\frac{\mathrm{12}}{{d}}−\mathrm{1}\right)=\frac{\mathrm{21}}{\mathrm{2}} \\ $$$$\mathrm{12}−{d}=\frac{\mathrm{21}}{\mathrm{2}}\Rightarrow{d}=\mathrm{12}−\frac{\mathrm{21}}{\mathrm{2}}=\frac{\mathrm{24}−\mathrm{21}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}. \\ $$$$\therefore{k}=\frac{\mathrm{6}}{\mathrm{3}/\mathrm{2}}=\mathrm{4}.\: \\ $$$${In}\:\left(\mathrm{2}\right)\:\Rightarrow\mathrm{4}\left({a}+\mathrm{4}×\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{30} \\ $$$$\mathrm{4}{a}+\mathrm{24}=\mathrm{30}\:{or}\:{a}=\frac{\mathrm{3}}{\mathrm{2}}. \\ $$$${The}\:{AP}\:{is}\:{hence}\:{the}\:{sequence}\:\left\{{u}\left({t}\right)\right\}_{{t}=\mathrm{1}} ^{\mathrm{8}} \\ $$$${where}\:{u}\left({t}\right)=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}+{t}−\mathrm{1}\right)=\frac{\mathrm{3}{t}}{\mathrm{2}},\:{and}\:{the} \\ $$$${number}\:{of}\:{terms}\:{is}\:\mathrm{2}{k}=\mathrm{2}×\mathrm{4}=\mathrm{8}. \\ $$$$ \\ $$
Answered by Yozzia last updated on 27/Aug/16
Check answer in comments.
$${Check}\:{answer}\:{in}\:{comments}. \\ $$
Answered by Rasheed Soomro last updated on 01/Sep/16
Technically Simple Approach Let a is first term and d is a common difference of the AP. The AP will be a,a+d,a+2d,a+3d,a+4d,...a+10(1/2) Let the number of terms is 2k a+(2k−1)d=a+10(1/2)⇒(2k−1)d=10(1/2)⇒d=((21)/(2(2k−1))) Now the AP will be a,a+(((21)/(2(2k−1)))),a+2(((21)/(2(2k−1)))),a+3(((21)/(2(2k−1)))),a+4(((21)/(2(2k−1)))),...a+10(1/2) The sum of odd terms a+(a+2(((21)/(2(2k−1)))))+(a+4(((21)/(2(2k−1)))))+...+(a+10(1/2)−(((21)/(2(2k−1))))).....k terms S=(n/2)(a+l) : a first term,l last term, n number of terms, S sum. S_o =(k/2)( a+(a−(((21)/(2(2k−1))))+10(1/2)) )=24 k( 2a+10(1/2)−(((21)/(2(2k−1)))))=48 2ak+10.5k−k(((21)/(2(2k−1))))=48................(i) Sum of even terms (a+(((21)/(2(2k−1)))))+(a+3(((21)/(2(2k−1)))))+(a+5(((21)/(2(2k−1)))))+...a+10(1/2)....k terms S_e =(k/2)( (a+((21)/(2(2k−1))))+(a+10(1/2)) )=30 k(2a+10(1/2)+(((21)/(2(2k−1)))))=60 2ak+10.5k+k(((21)/(2(2k−1))))=60................(ii) (ii)−(i): 2k(((21)/(2(2k−1))))=12⇒k(((21)/(2(2k−1))))=6⇒21k=12(2k−1) 24k−21k=12⇒k=4 (ii)+(i): k(4a+21)=108⇒4a+21=108/k=108/4=27 4a=6⇒a=(3/2) The first term of the AP is (3/2) and common difference is ((21)/(2(2k−1)))=((21)/(2(2(4)−1)))=(3/2) Hence The AP is (3/2),3,(9/2),6,((15)/2),9,((21)/2),12 ( 8 terms )
$${Technically}\:{Simple}\:{Approach} \\ $$$${Let}\:{a}\:{is}\:{first}\:{term}\:{and}\:{d}\:{is}\:{a}\:{common}\:{difference}\:{of}\:{the}\:{AP}. \\ $$$${The}\:{AP}\:{will}\:{be} \\ $$$$\:{a},{a}+{d},{a}+\mathrm{2}{d},{a}+\mathrm{3}{d},{a}+\mathrm{4}{d},…{a}+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Let}\:\:{the}\:{number}\:{of}\:{terms}\:{is}\:\mathrm{2}{k} \\ $$$${a}+\left(\mathrm{2}{k}−\mathrm{1}\right){d}={a}+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\left(\mathrm{2}{k}−\mathrm{1}\right){d}=\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{d}=\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)} \\ $$$${Now}\:{the}\:{AP}\:{will}\:{be} \\ $$$$\:{a},{a}+\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right),{a}+\mathrm{2}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right),{a}+\mathrm{3}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right),{a}+\mathrm{4}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right),…{a}+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${The}\:{sum}\:{of}\:{odd}\:{terms} \\ $$$${a}+\left({a}+\mathrm{2}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)\right)+\left({a}+\mathrm{4}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)\right)+…+\left({a}+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}−\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)\right)…..{k}\:{terms} \\ $$$${S}=\frac{{n}}{\mathrm{2}}\left({a}+{l}\right)\::\:{a}\:{first}\:{term},{l}\:{last}\:{term},\:{n}\:{number}\:{of}\:{terms},\:{S}\:{sum}. \\ $$$$\:\:\:\:\:\:\:\:{S}_{{o}} =\frac{{k}}{\mathrm{2}}\left(\:{a}+\left({a}−\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}\right)\:\right)=\mathrm{24} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{k}\left(\:\mathrm{2}{a}+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}−\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)\right)=\mathrm{48} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{ak}+\mathrm{10}.\mathrm{5}{k}−{k}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)=\mathrm{48}…………….\left({i}\right) \\ $$$${Sum}\:{of}\:{even}\:{terms} \\ $$$$\left({a}+\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)\right)+\left({a}+\mathrm{3}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)\right)+\left({a}+\mathrm{5}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)\right)+…{a}+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}….{k}\:{terms} \\ $$$$\:\:\:\:\:\:\:{S}_{{e}} =\frac{{k}}{\mathrm{2}}\left(\:\left({a}+\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)+\left({a}+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}\right)\:\right)=\mathrm{30} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{k}\left(\mathrm{2}{a}+\mathrm{10}\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)\right)=\mathrm{60} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{ak}+\mathrm{10}.\mathrm{5}{k}+{k}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)=\mathrm{60}…………….\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right):\:\:\:\mathrm{2}{k}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)=\mathrm{12}\Rightarrow{k}\left(\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}\right)=\mathrm{6}\Rightarrow\mathrm{21}{k}=\mathrm{12}\left(\mathrm{2}{k}−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{24}{k}−\mathrm{21}{k}=\mathrm{12}\Rightarrow{k}=\mathrm{4} \\ $$$$\left({ii}\right)+\left({i}\right):\:\:\:\:{k}\left(\mathrm{4}{a}+\mathrm{21}\right)=\mathrm{108}\Rightarrow\mathrm{4}{a}+\mathrm{21}=\mathrm{108}/{k}=\mathrm{108}/\mathrm{4}=\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{a}=\mathrm{6}\Rightarrow{a}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${The}\:{first}\:{term}\:\:{of}\:{the}\:{AP}\:\:{is}\:\:\frac{\mathrm{3}}{\mathrm{2}}\:{and}\: \\ $$$${common}\:{difference}\:{is}\:\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}{k}−\mathrm{1}\right)}=\frac{\mathrm{21}}{\mathrm{2}\left(\mathrm{2}\left(\mathrm{4}\right)−\mathrm{1}\right)}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${Hence}\:{The}\:{AP}\:{is} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}},\mathrm{3},\frac{\mathrm{9}}{\mathrm{2}},\mathrm{6},\frac{\mathrm{15}}{\mathrm{2}},\mathrm{9},\frac{\mathrm{21}}{\mathrm{2}},\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\left(\:\mathrm{8}\:{terms}\:\right) \\ $$

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