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.
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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If the first term of an AP is (-2) and the common difference is (6), what is the (12)th term?
B. (64)
\(a_{12}=-2+11\times6=64\)। पहले गुणा करें फिर (-2) जोड़ें। / \(a_{12}=-2+11\times6=64\). Multiply first and then add (-2).
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Find the (30)th term of the AP \(5,10,15,20,\ldots\).
B. (150)
यहाँ (a=5) और (d=5) है इसलिए \(a_{30}=5+29\times5=150\)। समान गुणकों वाली एपी में सूत्र जल्दी काम करता है। / Here (a=5) and (d=5), so \(a_{30}=5+29\times5=150\). The formula works quickly for APs with equal multiples.
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What is the (10)th term of the AP \(16,12,8,4,\ldots\)?
A. (-20)
यहाँ (d=-4) है इसलिए (a_{10}=16+9(-4)=-20)। घटते क्रम में उत्तर ऋणात्मक भी हो सकता है। / Here (d=-4), so (a_{10}=16+9(-4)=-20). In decreasing order, the answer can be negative.
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If (a=1) and (d=11), what is the (9)th term?
C. (89)
\(a_9=1+8\times11=89\)। (9)वें पद के लिए (8) अंतर जुड़ते हैं। / \(a_9=1+8\times11=89\). For the (9)th term, (8) differences are added.
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What is the (13)th term in the AP \(2,9,16,23,\ldots\)?
C. (86)
यहाँ (d=7) है इसलिए \(a_{13}=2+12\times7=86\)। सही पद संख्या के लिए (12d) जोड़ें। / Here (d=7), so \(a_{13}=2+12\times7=86\). Add (12d) for the correct term number.
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Find the (15)th term of the AP \(40,36,32,28,\ldots\).
C. (-16)
यहाँ (d=-4) है इसलिए (a_{15}=40+14(-4)=-16)। (14) बार सार्व अंतर जोड़ना है। / Here (d=-4), so (a_{15}=40+14(-4)=-16). The common difference is added (14) times.
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If the (5)th term of an AP is (22) and (d=6), what is the (8)th term?
D. (40)
(8)वाँ पद (5)वें पद से (3d) आगे है इसलिए (22+18=40)। पदों का अंतर गिनना आसान तरीका है। / The (8)th term is (3d) after the (5)th term, so (22+18=40). Counting the gap between terms is an easy method.
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If the (7)th term of an AP is (31) and (d=4), what is the (10)th term?
C. (43)
(10)वाँ पद (7)वें पद से (3d) आगे है इसलिए \(31+3\times4=43\)। पास के पदों के लिए अंतर विधि तेज होती है। / The (10)th term is (3d) after the (7)th term, so \(31+3\times4=43\). The difference method is quick for nearby terms.
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What is the (20)th term of the AP \(-5,-2,1,4,\ldots\)?
B. (52)
यहाँ (d=3) है इसलिए \(a_{20}=-5+19\times3=52\)। ऋणात्मक प्रथम पद को अंत में घटाएं। / Here (d=3), so \(a_{20}=-5+19\times3=52\). Subtract the negative first term effect carefully.
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Find the (8)th term in the AP \(18,23,28,33,\ldots\).
B. (53)
यहाँ (d=5) है इसलिए \(a_8=18+7\times5=53\)। (8)वें पद में (7) अंतर शामिल होते हैं। / Here (d=5), so \(a_8=18+7\times5=53\). The (8)th term includes (7) differences.
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What is the (11)th term of the AP \(8,15,22,29,\ldots\)?
B. (78)
यहाँ (d=7) है इसलिए \(a_{11}=8+10\times7=78\)। पहले पद से शुरू करके (10) अंतर जोड़ें। / Here (d=7), so \(a_{11}=8+10\times7=78\). Start from the first term and add (10) differences.
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If (a=25), (d=-2), and (n=17), what is \(a_n\)?
B. (-7)
(a_{17}=25+16(-2)=-7)। ऋणात्मक सार्व अंतर को सही से गुणा करें। / (a_{17}=25+16(-2)=-7). Multiply the negative common difference carefully.
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What is the (25)th term of the AP \(3,3,3,3,\ldots\)?
B. (3)
यहाँ (d=0) है इसलिए हर पद (3) रहेगा। समान पदों वाली एपी में \(a_n=a\) होता है। / Here (d=0), so every term remains (3). In a constant AP, \(a_n=a\).
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Find the (10)th term of the AP \(\frac{1}{2},1,\frac{3}{2},2,\ldots\).
C. (5)
यहाँ \(a=\frac{1}{2}\) और \(d=\frac{1}{2}\) है इसलिए \(a_{10}=5\)। भिन्नों में हर समान रखकर जोड़ें। / Here \(a=\frac{1}{2}\) and \(d=\frac{1}{2}\), so \(a_{10}=5\). Use a common denominator while adding fractions.
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What is the (21)st term of the AP \(0,5,10,15,\ldots\)?
B. (100)
यहाँ (a=0) और (d=5) है इसलिए \(a_{21}=0+20\times5=100\)। प्रथम पद (0) हो सकता है। / Here (a=0) and (d=5), so \(a_{21}=0+20\times5=100\). The first term can be (0).
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If the AP is \(4,9,14,19,\ldots\), what is the (16)th term?
C. (79)
यहाँ (d=5) है इसलिए \(a_{16}=4+15\times5=79\)। (16)वें पद के लिए (15d) जोड़ें। / Here (d=5), so \(a_{16}=4+15\times5=79\). For the (16)th term, add (15d).
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Find the (13)th term of the AP \(50,47,44,41,\ldots\).
B. (14)
यहाँ (d=-3) है इसलिए (a_{13}=50+12(-3)=14)। घटती एपी में पद छोटे होते जाते हैं। / Here (d=-3), so (a_{13}=50+12(-3)=14). In a decreasing AP, terms keep getting smaller.
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What is the (18)th term in the AP \(11,14,17,20,\ldots\)?
B. (62)
यहाँ (d=3) है इसलिए \(a_{18}=11+17\times3=62\)। लंबी पद संख्या में भी (n-1) का उपयोग करें। / Here (d=3), so \(a_{18}=11+17\times3=62\). Use (n-1) even for a larger term number.
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If (a=3) and (d=9), what is the (7)th term?
C. (57)
\(a_7=3+6\times9=57\)। (n) की जगह (7) रखने पर (6d) जुड़ता है। / \(a_7=3+6\times9=57\). When (n=7), (6d) is added.
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What is the (12)th term of the AP \(100,90,80,70,\ldots\)?
B. (-10)
यहाँ (d=-10) है इसलिए (a_{12}=100+11(-10)=-10)। बड़े अंतर में भी वही सूत्र लगता है। / Here (d=-10), so (a_{12}=100+11(-10)=-10). The same formula works for large differences too.
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Find the (10)th term of the AP \(6,13,20,27,\ldots\).
B. (69)
यहाँ (d=7) है इसलिए \(a_{10}=6+9\times7=69\)। (10)वें पद के लिए (9d) जोड़ें। / Here (d=7), so \(a_{10}=6+9\times7=69\). For the (10)th term, add (9d).
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If the first term of an AP is (15) and the common difference is (-4), what is the (8)th term?
A. (-13)
(a_8=15+7(-4)=-13)। ऋणात्मक (d) को कोष्ठक में लिखना उपयोगी है। / (a_8=15+7(-4)=-13). Writing negative (d) in brackets is useful.
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What is the (14)th term in the AP \(-10,-7,-4,-1,\ldots\)?
D. (29)
यहाँ (d=3) है इसलिए \(a_{14}=-10+13\times3=29\)। ऋणात्मक (a) के साथ जोड़ सही करें। / Here (d=3), so \(a_{14}=-10+13\times3=29\). Add carefully with a negative (a).
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What is the (9)th term of the AP \(13,17,21,25,\ldots\)?
C. (45)
यहाँ (d=4) है इसलिए \(a_9=13+8\times4=45\)। पहले दो पदों का अंतर जांचें। / Here (d=4), so \(a_9=13+8\times4=45\). Check the difference between the first two terms.
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If \(a_1=8\) and (d=2), what is the (20)th term?
B. (46)
\(a_{20}=8+19\times2=46\)। \(a_1\) ही प्रथम पद होता है। / \(a_{20}=8+19\times2=46\). \(a_1\) is the first term.
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Find the (6)th term of the AP \(30,25,20,15,\ldots\).
B. (5)
यहाँ (d=-5) है इसलिए (a_6=30+5(-5)=5)। घटते क्रम में सार्व अंतर ऋणात्मक लें। / Here (d=-5), so (a_6=30+5(-5)=5). Take the common difference as negative in decreasing order.
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What is the (15)th term of the AP \(1,4,7,10,\ldots\)?
C. (43)
यहाँ (a=1) और (d=3) है इसलिए \(a_{15}=1+14\times3=43\)। क्रमांक से (1) कम लेना जरूरी है। / Here (a=1) and (d=3), so \(a_{15}=1+14\times3=43\). Taking one less than the term number is necessary.
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If an AP has (a=4) and (d=7), what is its (5)th term?
C. (32)
\(a_5=4+4\times7=32\)। (5)वें पद के लिए (4d) जोड़ते हैं। / \(a_5=4+4\times7=32\). For the (5)th term, add (4d).
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What is the (11)th term in the AP \(9,6,3,0,\ldots\)?
B. (-21)
यहाँ (d=-3) है इसलिए (a_{11}=9+10(-3)=-21)। ऋण चिह्न को सही रखें। / Here (d=-3), so (a_{11}=9+10(-3)=-21). Keep the negative sign correct.
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