class10 se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
2084 questions tagged with class10.
If the (11)th term of an AP is (52) and (d=5), what is \(a_2\)?
A. (7)
\(a_2=a_{11}-9d=52-45=7\)। ज्ञात पद से पीछे आने पर स्थानों का अंतर गुणा करके घटाएं। / \(a_2=a_{11}-9d=52-45=7\). When moving backward from a known term, subtract the position gap times (d).
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What is the greatest term less than (100) in the AP \(1,4,7,\ldots\)?
C. (97)
इस AP के पद (1+3(n-1)) हैं। (100) से कम सबसे बड़ा ऐसा पद (97) है। / The terms are (1+3(n-1)). The greatest such term less than (100) is (97).
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In an AP, (a=30) and (d=-2). Which term will be (0)?
C. (16)वां(16)th
(0=30+(n-1)(-2)) से (2(n-1)=30), अतः (n=16)। शून्य पद के लिए भी वही (n)वां पद सूत्र लगाएं। / From (0=30+(n-1)(-2)), (2(n-1)=30), so (n=16). Use the same nth-term formula even for the zero term.
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If \(a_n=2n+5\), what are the first term and common difference of the AP?
A. (a=7,d=2)
(n=1) रखने पर \(a_1=7\) और लगातार पदों का अंतर (2) है। प्रत्यक्ष सूत्र में (d) (n) के गुणांक से मिलता है। / Putting (n=1), \(a_1=7\), and consecutive terms differ by (2). In a direct formula, (d) is the coefficient of (n).
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In the AP \(6,13,20,\ldots\), which is the last term just less than (90)?
A. (83)
(a_n=6+7(n-1))। (90) से कम पदों में (83) आता है और अगला (90) है। / (a_n=6+7(n-1)). Among terms less than (90), (83) occurs and the next term is (90).
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Which is the first negative term of the AP \(45,40,35,\ldots\)?
B. (11)वां(11)th
(a_n=45+(n-1)(-5)=50-5n)। \(a_n<0\) से (n>10), इसलिए पहला ऋणात्मक पद (11)वां है। / (a_n=45+(n-1)(-5)=50-5n). From \(a_n<0\), (n>10), so the first negative term is the (11)th.
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If the first term of an AP is (9) and \(a_{21}=69\), what is (d)?
B. (3)
(69=9+20d) से (60=20d), अतः (d=3)। (21)वें पद में (20d) जोड़ा जाता है। / From (69=9+20d), (60=20d), so (d=3). In the (21)st term, (20d) is added.
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If the (5)th term of an AP is (16) and the (9)th term is (32), what is the (13)th term?
C. (48)
\(d=\frac{32-16}{9-5}=4\), इसलिए \(a_{13}=32+4\times4=48\)। समान स्थान अंतर होने पर पदों का अंतर भी समान होता है। / \(d=\frac{32-16}{9-5}=4\), so \(a_{13}=32+4\times4=48\). Equal position gaps give equal term gaps in an AP.
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What will be the (19)th term of the AP \(-20,-14,-8,\ldots\)?
C. (88)
यहां (a=-20), (d=6), इसलिए \(a_{19}=-20+18\times6=88\)। ऋण से शुरू होने वाली बढ़ती AP में अंतिम जोड़ सावधानी से करें। / Here (a=-20), (d=6), so \(a_{19}=-20+18\times6=88\). In an increasing AP starting negative, add carefully at the end.
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Which term of the AP \(15,21,27,\ldots\) is (111)?
C. (17)वां(17)th
(111=15+(n-1)6), इसलिए (96=6(n-1)) और (n=17)। पद संख्या में अंतर को (d) से भाग दें। / (111=15+(n-1)6), so (96=6(n-1)) and (n=17). For term number, divide the difference by (d).
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The (8)th term of an AP is (35) and (d=4). What is \(a_1\)?
C. (7)
\(35=a+7\times4\), इसलिए (a=7)। (8)वें पद के लिए (7d) घटाएं। / \(35=a+7\times4\), so (a=7). For the (8)th term, subtract (7d).
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What is the (40)th term of the AP \(\frac{1}{2},1,\frac{3}{2},\ldots\)?
C. (20)
यहां \(a=\frac{1}{2}\), \(d=\frac{1}{2}\), इसलिए \(a_{40}=\frac{1}{2}+39\cdot\frac{1}{2}=20\)। भिन्नों को दशमलव में बदलकर भी जांच सकते हैं। / Here \(a=\frac{1}{2}\), \(d=\frac{1}{2}\), so \(a_{40}=\frac{1}{2}+39\cdot\frac{1}{2}=20\). Fractions can also be checked as decimals.
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If \(a_n=7-3n\), what is the (22)nd term of the AP?
A. (-59)
\(a_{22}=7-3\times22=-59\)। घटते प्रत्यक्ष सूत्र में ऋण चिह्न पर विशेष ध्यान दें। / \(a_{22}=7-3\times22=-59\). Pay special attention to the negative sign in a decreasing direct formula.
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If the (n)th term of an AP is given by \(a_n=4n-1\), what is \(a_{30}\)?
B. (119)
(n=30) रखने पर \(a_{30}=4\times30-1=119\)। प्रत्यक्ष सूत्र में केवल सही (n) रखें। / Putting (n=30), \(a_{30}=4\times30-1=119\). In a direct formula, substitute the correct (n).
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Find the (20)th term of the AP \(2.5,5,7.5,\ldots\).
B. (50)
यहां (a=2.5), (d=2.5), इसलिए \(a_{20}=2.5+19\times2.5=50\)। दशमलव AP में समान गुणन का ध्यान रखें। / Here (a=2.5), (d=2.5), so \(a_{20}=2.5+19\times2.5=50\). In decimal APs, keep multiplication exact.
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In an AP, \(a_3=11\) and (d=6). What is \(a_{17}\)?
B. (95)
(a_{17}=a_3+(17-3)d=11+14\times6=95)। किसी ज्ञात पद से सीधे आगे बढ़ना आसान है। / (a_{17}=a_3+(17-3)d=11+14\times6=95). Moving directly from a known term is easier.
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The first term of an AP is (18) and the (16)th term is (93). What is the common difference?
A. (5)
(93=18+15d), इसलिए (75=15d) और (d=5)। (16)वें पद के लिए गुणक (15) होगा। / (93=18+15d), so (75=15d) and (d=5). For the (16)th term, the multiplier is (15).
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What will be the (15)th term of the AP \(100,92,84,\ldots\)?
D. (-12)
यहां (d=-8), इसलिए (a_{15}=100+14(-8)=-12)। बड़े घटते अंतर में गुणा पहले करें। / Here (d=-8), so (a_{15}=100+14(-8)=-12). In large decreasing differences, multiply first.
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In an AP, \(a_4=14\) and \(a_{12}=46\). What is the value of \(a_{20}\)?
C. (78)
\(d=\frac{46-14}{12-4}=4\), इसलिए \(a_{20}=46+8\times4=78\)। पास वाले ज्ञात पद से आगे बढ़ना तेज तरीका है। / \(d=\frac{46-14}{12-4}=4\), so \(a_{20}=46+8\times4=78\). Moving from the nearest known term is a faster method.
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If the (10)th term of an AP is (41) and the (20)th term is (81), what is its common difference?
C. (4)
दो पदों का अंतर (81-41=40) है और पद संख्या का अंतर (10), इसलिए (d=4)। \(दो ज्ञात पदों में (d=\frac{\)पदों का अंतर}{स्थान का अंतर}) लगाएं। \(/ The difference of terms is (81-41=40) and the difference of positions is (10), so (d=4). Use (d=\frac{\)difference of terms}{difference of positions}).
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What is the (31)st term of the AP \(-4,1,6,\ldots\)?
B. (146)
यहां (a=-4), (d=5), इसलिए \(a_{31}=-4+30\times5=146\)। ऋणात्मक पहले पद को सावधानी से जोड़ें। / Here (a=-4), (d=5), so \(a_{31}=-4+30\times5=146\). Add the negative first term carefully.
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Which term of the AP \(3,8,13,\ldots\) is (88)?
A. (18)वां(18)th
(88=3+(n-1)5) से (85=5(n-1)), अतः (n=18)। पद संख्या के प्रश्न में दिए पद को \(a_n\) मानें। / From (88=3+(n-1)5), (85=5(n-1)), hence (n=18). In term-number questions, treat the given term as \(a_n\).
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The (6)th term of an AP is (23) and the common difference is (5). What is the first term?
D. (-2)
(23=a+5d=a+25), इसलिए (a=-2)। दिए गए पद से पीछे जाते समय (5d) घटाया जाता है। / (23=a+5d=a+25), so (a=-2). When moving backward from a given term, subtract (5d).
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In an AP, (a=5), (d=7), and \(a_n=103\). What is the value of (n)?
C. (15)
(103=5+(n-1)7) से (98=7(n-1)), अतः (n=15)। पद संख्या निकालते समय अंत में (1) जोड़ना न भूलें। / From (103=5+(n-1)7), (98=7(n-1)), so (n=15). When finding the term number, remember to add (1) at the end.
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Find the (25)th term of the AP \(12,9,6,\ldots\).
B. (-60)
यहां (a=12) और (d=-3), इसलिए (a_{25}=12+24(-3)=-60)। घटती श्रेणी में (d) का चिन्ह ध्यान रखें। / Here (a=12) and (d=-3), so (a_{25}=12+24(-3)=-60). Keep the sign of (d) carefully in decreasing APs.
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If the first term of an AP is (7) and the common difference is (4), what is its (18)th term?
A. (75)
सूत्र (a_n=a+(n-1)d) लगाने पर \(7+17\times4=75\)। परीक्षा में (n-1) को भूलना नहीं चाहिए। / Using (a_n=a+(n-1)d), \(7+17\times4=75\). Exam tip: do not forget (n-1).
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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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