100 results found for "easy amendment" in Class 10.
In which year was the Seventy-fourth Constitutional Amendment passed?
C. 19921992
चौहत्तरवां संशोधन 1992 में पारित हुआ था। इसे नगर निकायों और शहरी स्थानीय शासन से जोड़ें। / The Seventy-fourth Amendment was passed in 1992. Link it with municipalities and urban local governance.
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In which year was the Seventy-third Constitutional Amendment passed?
C. 19921992
तिहत्तरवां संशोधन 1992 में पारित हुआ था। इसे पंचायती राज और स्थानीय स्वशासन से जोड़ें। / The Seventy-third Amendment was passed in 1992. Link it with Panchayati Raj and local self-government.
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In which year was the Forty-second Constitutional Amendment passed?
C. 19761976
बयालीसवां संशोधन 1976 में पारित हुआ था। इसे संविधान में व्यापक बदलावों और प्रस्तावना में जोड़े गए शब्दों से जोड़ें। / The Forty-second Amendment was passed in 1976. Link it with wide constitutional changes and words added to the Preamble.
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The 52nd Constitutional Amendment related to anti defection law took place in which year?
C. 1985 ईस्वी1985 CE
52वां संविधान संशोधन 1985 में दल बदल विरोधी कानून से जुड़ा था। परीक्षा में प्रमुख संशोधनों और उनके विषय याद रखें। / The 52nd Constitutional Amendment of 1985 was linked with the anti defection law. For exams remember major amendments with their subjects.
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The 44th Constitutional Amendment was passed in which year?
C. 1978 ईस्वी1978 CE
44वां संविधान संशोधन 1978 में पारित हुआ था। परीक्षा में आपातकाल के बाद के संवैधानिक बदलाव याद रखें। / The 44th Constitutional Amendment was passed in 1978. For exams remember constitutional changes after the Emergency.
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The 73rd Constitutional Amendment came into effect in which year?
C. 1993 ईस्वी1993 CE
73वां संविधान संशोधन 1993 में लागू हुआ और पंचायतों को संवैधानिक दर्जा मिला। परीक्षा में स्थानीय स्वशासन से जुड़ी तिथियां याद रखें। / The 73rd Constitutional Amendment came into effect in 1993 and gave constitutional status to Panchayats. For exams remember dates related to local self government.
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The 73rd Amendment giving constitutional status to Panchayati Raj came into effect in which year?
C. 1993 ईस्वी1993 CE
73वां संशोधन 1993 में लागू हुआ और पंचायतों को संवैधानिक दर्जा मिला। परीक्षा में स्थानीय स्वशासन की तिथियां याद रखें। / The 73rd Amendment came into effect in 1993 and gave constitutional status to Panchayats. For exams remember dates of local self government.
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The 42nd Constitutional Amendment was passed in which year?
C. 1976 ईस्वी1976 CE
42वां संविधान संशोधन 1976 में पारित हुआ था। परीक्षा में प्रमुख संविधान संशोधनों के वर्ष याद रखें। / The 42nd Constitutional Amendment was passed in 1976. For exams remember years of major constitutional amendments.
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Susan B. Anthony's name is often linked with which U.S. constitutional amendment?
A. उन्नीसवां संशोधनNineteenth Amendment
उन्नीसवें संशोधन ने अमेरिकी महिलाओं को मतदान अधिकार दिया। परीक्षा में एंथनी को महिला मताधिकार से जोड़ें। / The Nineteenth Amendment gave American women voting rights. For exams connect Anthony with women's suffrage.
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Which line makes it easy to show height of a tree?
A. ऊर्ध्व रेखाVertical line
वृक्ष का तना ऊर्ध्व दिशा में बढ़ता है। परीक्षा में tall object के लिए vertical line देखें। / Tree trunk grows vertically. Exam tip: observe vertical line for tall object.
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In which situation is it easy to identify figure and background separately?
D. जब अच्छा विरोध होWhen there is good contrast
अच्छा विरोध आकृति और पृष्ठभूमि को अलग करता है। परीक्षा में figure ground relation में contrast याद रखें। / Good contrast separates figure and background. Exam tip: remember contrast in figure-ground relation.
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What is an easy difference between revolution and reform?
A. क्रांति बड़ा तेज बदलाव होती है सुधार क्रमिक बदलाव हो सकता हैRevolution is a major rapid change while reform can be gradual
क्रांति प्रायः गहरा बदलाव लाती है जबकि सुधार धीरे धीरे हो सकता है। परीक्षा में दोनों शब्दों को अलग समझें। / A revolution often brings deep change while reform can happen gradually. For exams understand the two terms separately.
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If (x+y=13) and (x=9), choose the correct value of (y).
C. (y=4)
(9+y=13), इसलिए (y=4)। दिए गए मान को रखने के बाद सरल घटाव करें। / (9+y=13), so (y=4). After placing the given value, do simple subtraction.
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In the equations (x+2y=14) and (x=4), what is (y)?
B. (y=5)
(x=4) रखने पर (4+2y=14), इसलिए (y=5)। जब एक चर दिया हो तो प्रतिस्थापन तुरंत करें। / Putting (x=4) gives (4+2y=14), so (y=5). When one variable is given, substitute immediately.
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Which easy method is most suitable to solve \(x^2+3x+2=0\)?
A. गुणनखंड विधिFactorisation method
यह ((x+1)(x+2)=0) में आसानी से टूटता है। परीक्षा में छोटे गुणांकों वाले प्रश्नों में गुणनखंड विधि तेज रहती है। / It easily factors as ((x+1)(x+2)=0). In exams, factorisation is fast for questions with small coefficients.
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Why was adopting khadi not always easy for poor people?
A. क्योंकि खादी कभी कभी मिल के कपड़े से महंगी होती थीBecause khadi was sometimes costlier than mill-made cloth
खादी स्वदेशी का आदर्श प्रतीक थी। / Khadi was an ideal symbol of swadeshi.
पर गरीब परिवारों के लिए कीमत बहुत महत्वपूर्ण थी। / But price was very important for poor families.
इससे आंदोलन के आदर्श और आर्थिक कठिनाई के बीच अंतर दिखता है। / This shows the gap between movement ideals and economic difficulty.
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Why was khadi not always an immediately easy option for poor people?
A. क्योंकि वह कई बार मिल के कपड़े से महंगी होती थीBecause it was often costlier than mill-made cloth
खादी स्वदेशी का प्रतीक थी। / Khadi was a symbol of swadeshi.
पर गरीबों के लिए कीमत महत्वपूर्ण होती है। / But price mattered a lot for poor people.
इसलिए आंदोलन के आदर्श और आर्थिक कठिनाई में अंतर समझना चाहिए। / Therefore one must understand the gap between movement ideals and economic difficulty.
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What is an easy way to identify the difference between Marianne and Germania?
C. मैरिआन फ्रांस से और जर्मानिया जर्मनी से जुड़ी हैMarianne is linked with France and Germania with Germany
दोनों राष्ट्र के महिला रूपक हैं। / Both are female allegories of the nation.
मैरिआन फ्रांस की और जर्मानिया जर्मनी की प्रतीक है। / Marianne symbolises France and Germania symbolises Germany.
परीक्षा में देश के आधार पर दोनों को अलग पहचानें। / In exams identify them by their countries.
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What is the easy identification of reduction?
B. ऑक्सीजन का हटनाRemoval of oxygen
अपचयन में सामान्यतः ऑक्सीजन हटती है। / Reduction generally involves removal of oxygen.
जिस पदार्थ से ऑक्सीजन हटती है वह अपचयित होता है। / The substance losing oxygen is reduced.
ऑक्सीकरण और अपचयन को विपरीत रूप में याद रखें। / Remember oxidation and reduction as opposite processes.
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What is the easy identification of oxidation?
A. ऑक्सीजन का जुड़नाAddition of oxygen
ऑक्सीकरण में सामान्यतः ऑक्सीजन जुड़ती है। / Oxidation generally involves addition of oxygen.
जिस पदार्थ में ऑक्सीजन जुड़ती है वह ऑक्सीकृत होता है। / The substance gaining oxygen is oxidised.
आसान प्रश्नों में ऑक्सीजन की दिशा देखें। / In easy questions check the direction of oxygen.
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Find the sum of the first (14) terms of the arithmetic progression \(22,25,28,\ldots\).
B. (581)
चौदहवाँ पद (61) है, इसलिए (S_{14}=\frac{14}{2}(22+61)=581)। अंतिम पद की सही गणना से योग सही मिलता है। / The fourteenth term is (61), so (S_{14}=\frac{14}{2}(22+61)=581). Correct calculation of the last term gives the correct sum.
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What is the sum of the first (16) odd natural numbers?
B. (256)
पहले (n) विषम संख्याओं का योग \(n^2\) होता है, इसलिए \(16^2=256\)। यह सूत्र याद रखने योग्य है। / The sum of the first (n) odd numbers is \(n^2\), so \(16^2=256\). This formula is worth remembering.
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If (S_n=\frac{n}{2}(a+l)), (a=14), (l=84), and (n=15), find the sum.
C. (735)
(S_{15}=\frac{15}{2}(14+84)=735)। पहले और अंतिम पद का औसत लेकर (n) से गुणा करें। / (S_{15}=\frac{15}{2}(14+84)=735). Take the average of the first and last terms and multiply by (n).
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B. (177)
छठे से ग्यारहवें पदों का योग \(S_{11}-S_5=177\) है। आंशिक योगों का अंतर सीधे उत्तर देता है। / The sum of the (6)th to (11)th terms is \(S_{11}-S_5=177\). The difference of partial sums gives the answer directly.
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What is the sum of the first (13) terms of the arithmetic progression \(2,9,16,\ldots\)?
D. (572)
तेरहवाँ पद (86) है, इसलिए (S_{13}=\frac{13}{2}(2+86)=572)। अंतिम पद निकालते समय ((n-1)d) का उपयोग करें। / The thirteenth term is (86), so (S_{13}=\frac{13}{2}(2+86)=572). Use ((n-1)d) when finding the last term.
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If an arithmetic progression has (a=20), (d=-3), and (n=7), what is the sum of the first (7) terms?
A. (77)
सातवाँ पद (2) है और (S_7=\frac{7}{2}(20+2)=77)। ऋणात्मक (d) में चिन्ह की गलती न करें। / The seventh term is (2), and (S_7=\frac{7}{2}(20+2)=77). Do not make a sign error with negative (d).
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What will be the sum of the first (19) even natural numbers?
B. (380)
पहले (n) सम संख्याओं का योग (n(n+1)) होता है, इसलिए \(19\times20=380\)। सम संख्याएँ \(2,4,6,\ldots\) से शुरू होती हैं। / The sum of the first (n) even numbers is (n(n+1)), so \(19\times20=380\). Even numbers start from \(2,4,6,\ldots\).
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What is the sum of the first (22) natural numbers?
B. (253)
पहले (n) प्राकृतिक संख्याओं का योग (\frac{n(n+1)}{2}) होता है, इसलिए (253) मिलेगा। ऐसे प्रश्न में सीधे (n) का मान लगाएँ। / The sum of the first (n) natural numbers is (\frac{n(n+1)}{2}), so the answer is (253). Put the value of (n) directly in such questions.
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Find the sum of the first (14) terms of the arithmetic progression \(2,5,8,\ldots\).
C. (301)
चौदहवाँ पद (41) है, इसलिए (S_{14}=\frac{14}{2}(2+41)=301)। अंतिम पद सही निकालना मुख्य कदम है। / The fourteenth term is (41), so (S_{14}=\frac{14}{2}(2+41)=301). Finding the last term correctly is the main step.
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Find the sum of the first (10) terms of the arithmetic progression \(14,17,20,\ldots\).
B. (275)
दसवाँ पद (41) है और (S_{10}=\frac{10}{2}(14+41)=275)। अंतिम पद निकालकर योग लेना सरल है। / The tenth term is (41), and (S_{10}=\frac{10}{2}(14+41)=275). Finding the last term first makes the sum simple.
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Find the sum of the first (15) terms of the arithmetic progression \(9,13,17,\ldots\).
A. (555)
अंतिम पद (65) है, इसलिए (S_{15}=\frac{15}{2}(9+65)=555)। अंतिम पद से गणना आसान हो सकती है। / The last term is (65), so (S_{15}=\frac{15}{2}(9+65)=555). Using the last term can simplify calculation.
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C. (225)
योग (=) औसत \(\times\) पदों की संख्या, इसलिए \(25\times9=225\)। औसत दिए होने पर लंबा सूत्र जरूरी नहीं। / Sum equals average \(\times\) number of terms, so \(25\times9=225\). When the average is given, the long formula is not needed.
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What is the sum of the first (10) terms of the AP \(17,22,27,\ldots\)?
C. (395)
दसवां पद (62) है। (S_{10}=\frac{10}{2}(17+62)=395)। / The tenth term is (62). (S_{10}=\frac{10}{2}(17+62)=395).
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An AP has (a=2), (l=38), and (n=10). What is \(S_n\)?
C. (200)
पहला और अंतिम पद दिए हैं। (S_{10}=\frac{10}{2}(2+38)=200)। / The first and last terms are given. (S_{10}=\frac{10}{2}(2+38)=200).
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What is the sum of the first (10) terms of the AP \(2,5,8,\ldots\)?
C. (155)
यहां (a=2), (d=3), (n=10)। \(S_{10}=\frac{10}{2}[2\cdot2+9\cdot3]=155\)। / Here (a=2), (d=3), (n=10). \(S_{10}=\frac{10}{2}[2\cdot2+9\cdot3]=155\).
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Find the (6)th term of the AP \(\frac{4}{5},\frac{9}{5},\frac{14}{5},\frac{19}{5},\ldots\).
B. \(\frac{29}{5}\)
यहाँ (d=1) है इसलिए \(a_6=\frac{4}{5}+5=\frac{29}{5}\)। पूर्ण संख्या को समान हर वाली भिन्न में बदलें। / Here (d=1), so \(a_6=\frac{4}{5}+5=\frac{29}{5}\). Convert the whole number to a fraction with the same denominator.
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What is the (40)th term of the AP \(14,14,14,14,\ldots\)?
B. (14)
यहाँ (d=0) है इसलिए हर पद (14) रहेगा। स्थिर एपी में \(a_n=a\) होता है। / Here (d=0), so every term remains (14). In a constant AP, \(a_n=a\).
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If \(a_9=70\) and (d=3), what is the (15)th term?
C. (88)
(15)वाँ पद (9)वें पद से (6d) आगे है इसलिए (70+18=88)। निकट पदों में यह विधि सरल है। / The (15)th term is (6d) after the (9)th term, so (70+18=88). This method is simple for nearby terms.
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What is the (28)th term of the AP \(9,15,21,27,\ldots\)?
B. (171)
यहाँ (d=6) है इसलिए \(a_{28}=9+27\times6=171\)। (28)वें पद के लिए (27d) जोड़ें। / Here (d=6), so \(a_{28}=9+27\times6=171\). For the (28)th term, add (27d).
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If \(a_1=22\) and (d=13), what is the (5)th term?
C. (74)
\(a_5=22+4\times13=74\)। \(a_1\) को प्रथम पद माना जाता है। / \(a_5=22+4\times13=74\). \(a_1\) is treated as the first term.
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Find the (14)th term of the AP \(33,29,25,21,\ldots\).
B. (-19)
यहाँ (d=-4) है इसलिए (a_{14}=33+13(-4)=-19)। (13) बार (-4) जोड़ें। / Here (d=-4), so (a_{14}=33+13(-4)=-19). Add (-4) thirteen times.
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What is the (16)th term of the AP \(6,17,28,39,\ldots\)?
B. (171)
यहाँ (d=11) है इसलिए \(a_{16}=6+15\times11=171\)। (16)वें पद के लिए (15) अंतर जोड़ें। / Here (d=11), so \(a_{16}=6+15\times11=171\). For the (16)th term, add (15) differences.
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If (a=120), (d=-9), and (n=13), find \(a_n\).
B. (12)
(a_{13}=120+12(-9)=12)। पहले \(12\times9\) घटाएं। / (a_{13}=120+12(-9)=12). First subtract \(12\times9\).
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What is the (18)th term of the AP \(-15,-9,-3,3,\ldots\)?
C. (87)
यहाँ (d=6) है इसलिए \(a_{18}=-15+17\times6=87\)। ऋणात्मक प्रथम पद को सही जोड़ें। / Here (d=6), so \(a_{18}=-15+17\times6=87\). Add the negative first term correctly.
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Find the (24)th term of the AP \(12,17,22,27,\ldots\).
B. (127)
यहाँ (d=5) है इसलिए \(a_{24}=12+23\times5=127\)। (24)वें पद के लिए (23d) जोड़ें। / Here (d=5), so \(a_{24}=12+23\times5=127\). For the (24)th term, add (23d).
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If \(a_8=44\) and (d=-2), what is the (12)th term?
B. (36)
(12)वाँ पद (8)वें पद से (4d) आगे है इसलिए (44+4(-2)=36)। ऋणात्मक अंतर को ध्यान से जोड़ें। / The (12)th term is (4d) after the (8)th term, so (44+4(-2)=36). Add the negative difference carefully.
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What is the (21)st term of the AP \(7,16,25,34,\ldots\)?
B. (187)
यहाँ (d=9) है इसलिए \(a_{21}=7+20\times9=187\)। (21)वें पद तक (20) अंतर जुड़ते हैं। / Here (d=9), so \(a_{21}=7+20\times9=187\). Up to the (21)st term, (20) differences are added.
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If an AP has (a=2.5), (d=2.5), and (n=10), what is \(a_n\)?
B. (25.0)
\(a_{10}=2.5+9\times2.5=25.0\)। दशमलव को सामान्य संख्या की तरह रखें। / \(a_{10}=2.5+9\times2.5=25.0\). Treat decimals like ordinary numbers.
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Find the (11)th term of the AP \(48,42,36,30,\ldots\).
B. (-12)
यहाँ (d=-6) है इसलिए (a_{11}=48+10(-6)=-12)। (11)वें पद के लिए (10d) जोड़ें। / Here (d=-6), so (a_{11}=48+10(-6)=-12). For the (11)th term, add (10d).
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If \(a_2=12\) and (d=9), what is the (7)th term?
B. (57)
(7)वाँ पद (2)रे पद से (5d) आगे है इसलिए (12+45=57)। दिए पद से आगे बढ़कर हल करें। / The (7)th term is (5d) after the (2)nd term, so (12+45=57). Solve by moving forward from the given term.
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What is the (9)th term of the AP \(18,28,38,48,\ldots\)?
C. (98)
यहाँ (d=10) है इसलिए \(a_9=18+8\times10=98\)। (9)वें पद तक (8) अंतर जुड़ते हैं। / Here (d=10), so \(a_9=18+8\times10=98\). Up to the (9)th term, (8) differences are added.
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If (a=5) and (d=-1), what is the (30)th term?
C. (-24)
(a_{30}=5+29(-1)=-24)। (30)वें पद के लिए (29d) जोड़ना है। / (a_{30}=5+29(-1)=-24). For the (30)th term, add (29d).
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What is the (22)nd term of the AP \(31,34,37,40,\ldots\)?
D. (94)
यहाँ (d=3) है इसलिए \(a_{22}=31+21\times3=94\)। (22)वें पद में (21) अंतर जुड़ते हैं। / Here (d=3), so \(a_{22}=31+21\times3=94\). The (22)nd term includes (21) differences.
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Find the (9)th term of the AP \(\frac{5}{4},\frac{7}{4},\frac{9}{4},\frac{11}{4},\ldots\).
C. \(\frac{21}{4}\)
यहाँ \(d=\frac{1}{2}\) है इसलिए \(a_9=\frac{5}{4}+8\times\frac{1}{2}=\frac{21}{4}\)। भिन्न वाले अंतर को सरल रूप में लें। / Here \(d=\frac{1}{2}\), so \(a_9=\frac{5}{4}+8\times\frac{1}{2}=\frac{21}{4}\). Use the fractional difference carefully.
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If (a=0), (d=11), and (n=11), what is \(a_n\)?
B. (110)
\(a_{11}=0+10\times11=110\)। प्रथम पद (0) होने पर भी (n-1) ही लेते हैं। / \(a_{11}=0+10\times11=110\). Even when the first term is (0), use (n-1).
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What is the (16)th term of the AP \(200,190,180,170,\ldots\)?
B. (50)
यहाँ (d=-10) है इसलिए (a_{16}=200+15(-10)=50)। बड़े पदों में भी चिह्न पर ध्यान दें। / Here (d=-10), so (a_{16}=200+15(-10)=50). Watch the sign even with large terms.
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If the (4)th term of an AP is (21) and (d=6), what is the (10)th term?
C. (57)
(10)वाँ पद (4)थे पद से (6d) आगे है इसलिए (21+36=57)। पद संख्या का अंतर सीधे उपयोग करें। / The (10)th term is (6d) after the (4)th term, so (21+36=57). Use the difference in term numbers directly.
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Find the (14)th term of the AP \(5,14,23,32,\ldots\).
C. (122)
यहाँ (d=9) है इसलिए \(a_{14}=5+13\times9=122\)। (n-1=13) का उपयोग करें। / Here (d=9), so \(a_{14}=5+13\times9=122\). Use (n-1=13).
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What is the (18)th term of the AP \(64,60,56,52,\ldots\)?
A. (-4)
यहाँ (d=-4) है इसलिए (a_{18}=64+17(-4)=-4)। (17) बार ऋणात्मक अंतर जोड़ें। / Here (d=-4), so (a_{18}=64+17(-4)=-4). Add the negative difference (17) times.
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If \(a_6=38\) and (d=5), what is the (13)th term?
C. (73)
(13)वाँ पद (6)वें पद से (7d) आगे है इसलिए (38+35=73)। दिए पद से आगे का अंतर सही गिनें। / The (13)th term is (7d) after the (6)th term, so (38+35=73). Count the forward gap correctly.
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Find the (8)th term of the AP \(1.5,3.0,4.5,6.0,\ldots\).
D. (12.0)
यहाँ (a=1.5) और (d=1.5) है इसलिए \(a_8=12.0\)। दशमलव में भी वही सूत्र लागू होता है। / Here (a=1.5) and (d=1.5), so \(a_8=12.0\). The same formula applies to decimals too.
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What is the (19)th term of the AP \(11,19,27,35,\ldots\)?
C. (155)
यहाँ (d=8) है इसलिए \(a_{19}=11+18\times8=155\)। (19)वें पद के लिए (18d) जोड़ें। / Here (d=8), so \(a_{19}=11+18\times8=155\). For the (19)th term, add (18d).
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If (a=18) and (d=-7), what is the (6)th term?
A. (-17)
(a_6=18+5(-7)=-17)। घटती एपी में उत्तर ऋणात्मक भी हो सकता है। / (a_6=18+5(-7)=-17). In a decreasing AP, the answer can be negative.
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What is the (15)th term in the AP \(-20,-15,-10,-5,\ldots\)?
C. (50)
यहाँ (d=5) है इसलिए \(a_{15}=-20+14\times5=50\)। ऋणात्मक शुरुआत से भ्रमित न हों। / Here (d=5), so \(a_{15}=-20+14\times5=50\). Do not get confused by the negative start.
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Find the (12)th term of the AP \(13,23,33,43,\ldots\).
B. (123)
यहाँ (d=10) है इसलिए \(a_{12}=13+11\times10=123\)। (12)वें पद में (11) सार्व अंतर जुड़ते हैं। / Here (d=10), so \(a_{12}=13+11\times10=123\). The (12)th term includes (11) common differences.
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If \(a_3=16\) and (d=7), what is the (8)th term?
B. (51)
(8)वाँ पद (3)रे पद से (5d) आगे है इसलिए (16+35=51)। पदों के बीच का अंतर गिनें। / The (8)th term is (5d) after the (3)rd term, so (16+35=51). Count the gap between terms.
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What is the (25)th term of the AP \(100,97,94,91,\ldots\)?
A. (28)
यहाँ (d=-3) है इसलिए (a_{25}=100+24(-3)=28)। (25)वें पद तक (24) अंतर जुड़ते हैं। / Here (d=-3), so (a_{25}=100+24(-3)=28). Up to the (25)th term, (24) differences are added.
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If (a=7), (d=12), and (n=8), what is \(a_n\)?
B. (91)
\(a_8=7+7\times12=91\)। (n=8) होने पर (7d) जुड़ता है। / \(a_8=7+7\times12=91\). When (n=8), (7d) is added.
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Find the (20)th term of the AP \(55,52,49,46,\ldots\).
B. (-2)
यहाँ (d=-3) है इसलिए (a_{20}=55+19(-3)=-2)। (20)वें पद के लिए (19d) जोड़ें। / Here (d=-3), so (a_{20}=55+19(-3)=-2). For the (20)th term, add (19d).
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What is the (10)th term of the AP \(25,31,37,43,\ldots\)?
D. (79)
यहाँ (d=6) है इसलिए \(a_{10}=25+9\times6=79\)। (10)वें पद में (9) अंतर होते हैं। / Here (d=6), so \(a_{10}=25+9\times6=79\). The (10)th term has (9) differences.
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If the first term of an AP is (-12) and the common difference is (5), what is the (16)th term?
B. (63)
\(a_{16}=-12+15\times5=63\)। पहले (15d) निकालें फिर प्रथम पद जोड़ें। / \(a_{16}=-12+15\times5=63\). First calculate (15d), then add the first term.
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Find the (18)th term of the AP \(2,11,20,29,\ldots\).
C. (155)
यहाँ (d=9) है इसलिए \(a_{18}=2+17\times9=155\)। बड़ी पद संख्या में भी वही सूत्र लगाएं। / Here (d=9), so \(a_{18}=2+17\times9=155\). Use the same formula even for a large term number.
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What is the (9)th term of the AP \(90,82,74,66,\ldots\)?
B. (26)
यहाँ (d=-8) है इसलिए (a_9=90+8(-8)=26)। सार्व अंतर का चिह्न सही रखें। / Here (d=-8), so (a_9=90+8(-8)=26). Keep the sign of the common difference correct.
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If \(a_5=27\) and (d=4), what is the (11)th term?
C. (51)
(11)वाँ पद (5)वें पद से (6d) आगे है इसलिए (27+24=51)। दिए पद से आगे बढ़ना तेज तरीका है। / The (11)th term is (6d) after the (5)th term, so (27+24=51). Moving ahead from the given term is quick.
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Find the (12)th term of the AP \(\frac{1}{3},\frac{2}{3},1,\frac{4}{3},\ldots\).
D. (4)
यहाँ \(a=\frac{1}{3}\) और \(d=\frac{1}{3}\) है इसलिए \(a_{12}=4\)। भिन्नों में हर समान रखें। / Here \(a=\frac{1}{3}\) and \(d=\frac{1}{3}\), so \(a_{12}=4\). Keep denominators common in fractions.
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What is the (17)th term of the AP \(15,22,29,36,\ldots\)?
B. (127)
यहाँ (d=7) है इसलिए \(a_{17}=15+16\times7=127\)। (17)वें पद के लिए (16) अंतर जोड़ें। / Here (d=7), so \(a_{17}=15+16\times7=127\). For the (17)th term, add (16) differences.
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If (a=80), (d=-5), and (n=14), find \(a_n\).
B. (15)
(a_{14}=80+13(-5)=15)। ऋणात्मक (d) को कोष्ठक में लिखें। / (a_{14}=80+13(-5)=15). Write negative (d) in brackets.
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What is the (15)th term of the AP \(4,12,20,28,\ldots\)?
C. (116)
यहाँ (a=4) और (d=8) है इसलिए \(a_{15}=4+14\times8=116\)। (n-1) को सही रखें। / Here (a=4) and (d=8), so \(a_{15}=4+14\times8=116\). Keep (n-1) correct.
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If the first term of an AP is (9) and the common difference is (10), what is the (7)th term?
B. (69)
\(a_7=9+6\times10=69\)। (7)वें पद के लिए (6d) जोड़ना होता है। / \(a_7=9+6\times10=69\). For the (7)th term, (6d) is added.
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What is the (8)th term in the AP \(72,66,60,54,\ldots\)?
C. (30)
यहाँ (d=-6) है इसलिए (a_8=72+7(-6)=30)। (8)वें पद के लिए (7d) जोड़ें। / Here (d=-6), so (a_8=72+7(-6)=30). For the (8)th term, add (7d).
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Find the (13)th term of the AP \(-4,2,8,14,\ldots\).
C. (68)
यहाँ (d=6) है इसलिए \(a_{13}=-4+12\times6=68\)। ऋणात्मक प्रथम पद के साथ जोड़ सावधानी से करें। / Here (d=6), so \(a_{13}=-4+12\times6=68\). Add carefully when the first term is negative.
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If (a=6) and (d=8), what is the (12)th term?
C. (94)
\(a_{12}=6+11\times8=94\)। (12)वें पद तक (11) अंतर जुड़ते हैं। / \(a_{12}=6+11\times8=94\). Up to the (12)th term, (11) differences are added.
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What is the (10)th term of the AP \(45,41,37,33,\ldots\)?
A. (9)
यहाँ (d=-4) है इसलिए (a_{10}=45+9(-4)=9)। घटती एपी में (d) ऋणात्मक लें। / Here (d=-4), so (a_{10}=45+9(-4)=9). In a decreasing AP, take (d) as negative.
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Find the (9)th term of the AP \(3,8,13,18,\ldots\).
C. (43)
यहाँ (a=3) और (d=5) है इसलिए \(a_9=3+8\times5=43\)। पहले सार्व अंतर पहचानें। / Here (a=3) and (d=5), so \(a_9=3+8\times5=43\). First identify the common difference.
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If an AP has (a=14), (d=3), and (n=11), what is \(a_n\)?
B. (44)
(a_n=a+(n-1)d) से \(a_{11}=14+10\times3=44\)। परीक्षा में पहले (n-1) निकालें। / Using (a_n=a+(n-1)d), \(a_{11}=14+10\times3=44\). In exams, first find (n-1).
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Find the (7)th term of the AP \(\frac{2}{3},\frac{5}{3},\frac{8}{3},\frac{11}{3},\ldots\).
C. \(\frac{20}{3}\)
यहाँ (d=1) है इसलिए \(a_7=\frac{2}{3}+6=\frac{20}{3}\)। पूर्ण संख्या को भिन्न में बदलकर जोड़ें। / Here (d=1), so \(a_7=\frac{2}{3}+6=\frac{20}{3}\). Convert the whole number into a fraction before adding.
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What is the (100)th term of the AP \(6,6,6,6,\ldots\)?
A. (6)
यहाँ (d=0) है इसलिए \(a_{100}=6\)। स्थिर एपी में हर पद समान होता है। / Here (d=0), so \(a_{100}=6\). In a constant AP, every term is the same.
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If \(a_2=10\) and (d=4), what is the (6)th term?
C. (26)
(6)वाँ पद (2)रे पद से (4d) आगे है इसलिए (10+16=26)। दिए पद से आगे बढ़कर हल करें। / The (6)th term is (4d) after the (2)nd term, so (10+16=26). Solve by moving forward from the given term.
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Find the (18)th term of the AP \(35,30,25,20,\ldots\).
A. (-50)
यहाँ (d=-5) है इसलिए (a_{18}=35+17(-5)=-50)। (17) बार (-5) जोड़ना है। / Here (d=-5), so (a_{18}=35+17(-5)=-50). You need to add (-5) seventeen times.
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What is the (22)nd term of the AP \(1,8,15,22,\ldots\)?
B. (148)
यहाँ (d=7) है इसलिए \(a_{22}=1+21\times7=148\)। (22)वें पद के लिए (21d) जोड़ें। / Here (d=7), so \(a_{22}=1+21\times7=148\). For the (22)nd term, add (21d).
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If (a=60) and (d=-6), what is the (9)th term?
B. (12)
(a_9=60+8(-6)=12)। पहले (8d) निकालें फिर प्रथम पद में जोड़ें। / (a_9=60+8(-6)=12). First calculate (8d), then add it to the first term.
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Find the (11)th term of the AP \(-8,-3,2,7,\ldots\).
B. (42)
यहाँ (d=5) है इसलिए \(a_{11}=-8+10\times5=42\)। ऋणात्मक शुरुआत के बाद भी सूत्र वही रहता है। / Here (d=5), so \(a_{11}=-8+10\times5=42\). The formula remains the same even with a negative start.
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What is the (6)th term of the AP \(17,24,31,38,\ldots\)?
B. (52)
यहाँ (d=7) है इसलिए \(a_6=17+5\times7=52\)। (6)वें पद में (5) अंतर होते हैं। / Here (d=7), so \(a_6=17+5\times7=52\). The (6)th term has (5) differences.
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If (a=9), (d=6), and (n=13), find \(a_n\).
C. (81)
\(a_{13}=9+12\times6=81\)। (n-1=12) का उपयोग करें। / \(a_{13}=9+12\times6=81\). Use (n-1=12).
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What is the (19)th term of the AP \(24,21,18,15,\ldots\)?
A. (-30)
यहाँ (d=-3) है इसलिए (a_{19}=24+18(-3)=-30)। (19)वें पद के लिए (18d) जुड़ता है। / Here (d=-3), so (a_{19}=24+18(-3)=-30). For the (19)th term, (18d) is added.
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If the (3)rd term of an AP is (14) and (d=5), what is the (7)th term?
C. (34)
(7)वाँ पद (3)रे पद से (4d) आगे है इसलिए (14+20=34)। बीच के पदों की संख्या सही गिनें। / The (7)th term is (4d) after the (3)rd term, so (14+20=34). Count the gap between terms correctly.
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What is the (8)th term of the AP \(\frac{3}{2},\frac{5}{2},\frac{7}{2},\frac{9}{2},\ldots\)?
B. \(\frac{17}{2}\)
यहाँ \(a=\frac{3}{2}\) और (d=1) है इसलिए \(a_8=\frac{17}{2}\)। भिन्न वाले पदों में हर को ध्यान से देखें। / Here \(a=\frac{3}{2}\) and (d=1), so \(a_8=\frac{17}{2}\). Watch the denominator carefully in fractional terms.
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Find the (14)th term of the AP \(12,20,28,36,\ldots\).
C. (116)
यहाँ (d=8) है इसलिए \(a_{14}=12+13\times8=116\)। (13d) जोड़ना सही है। / Here (d=8), so \(a_{14}=12+13\times8=116\). Adding (13d) is correct.
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If \(a_4=18\) and (d=3), what is the (9)th term?
D. (33)
(9)वाँ पद (4)वें पद से (5d) आगे है इसलिए (18+15=33)। दिए हुए पद से आगे बढ़ना तेज तरीका है। / The (9)th term is (5d) after the (4)th term, so (18+15=33). Moving forward from the given term is quick.
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What is the (16)th term of the AP \(21,19,17,15,\ldots\)?
A. (-9)
यहाँ (d=-2) है इसलिए (a_{16}=21+15(-2)=-9)। (16)वें पद तक (15) बार अंतर जुड़ता है। / Here (d=-2), so (a_{16}=21+15(-2)=-9). Up to the (16)th term, the difference is added (15) times.
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