nth_term se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
254 questions tagged with nth_term.
What is the (28)th term of the AP \(4,11,18,\ldots\)?
B. (193)
(a=4) और (d=7) हैं इसलिए \(a_{28}=4+27\times7=193\)। परीक्षा में पद संख्या से (1) घटाना न भूलें। / (a=4) and (d=7) so \(a_{28}=4+27\times7=193\). In exams remember to subtract (1) from the term number.
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In the AP \(64,58,52,\ldots\), which is the last term greater than (10)?
C. (16)
पद \(64,58,52,\ldots\) में (d=-6) है। (10) से बड़ा अंतिम पद (16) है क्योंकि अगला पद (10) है। / The terms \(64,58,52,\ldots\) have (d=-6). The last term greater than (10) is (16) because the next term is (10).
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If in an AP \(a_5+a_9=60\) and (d=3), what is \(a_7\)?
B. (30)
AP में समान दूरी वाले पदों का औसत बीच वाला पद होता है, इसलिए \(a_7=\frac{60}{2}=30\)। सममित पदों में औसत विधि तेज होती है। / In an AP, the average of equally spaced terms is the middle term, so \(a_7=\frac{60}{2}=30\). For symmetric terms, the average method is faster.
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In an AP, \(a_1+a_2=18\) and (d=4). What is \(a_{10}\)?
B. (43)
(a_1+a_2=a+(a+4)=18), इसलिए (a=7)। अब \(a_{10}=7+9\times4=43\)। पहले पहला पद निकालें, फिर (n)वां पद। / (a_1+a_2=a+(a+4)=18), so (a=7). Now \(a_{10}=7+9\times4=43\). First find the first term, then the nth term.
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If \(a_{n+1}-a_n=8\) and \(a_4=30\), what is \(a_{16}\)?
C. (126)
यहां (d=8), इसलिए \(a_{16}=a_4+12d=30+96=126\)। \(a_{n+1}-a_n\) ही AP का सार्व अंतर होता है। / Here (d=8), so \(a_{16}=a_4+12d=30+96=126\). \(a_{n+1}-a_n\) is the common difference of the AP.
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The (n)th term of the AP \(11,17,23,\ldots\) is \(a_n=6n+5\). What is its (32)nd term?
B. (197)
\(a_{32}=6\times32+5=197\)। बनाए गए सूत्र में (n) का सही मान रखें। / \(a_{32}=6\times32+5=197\). Substitute the correct value of (n) in the formed formula.
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What is the last term among the positive multiples of (9) less than (200)?
C. (198)
\(9,18,27,\ldots\) में (9n<200), इसलिए सबसे बड़ा (n=22) और पद (198) है। गुणजों में सीमा से कम सबसे बड़ा पूर्णांक लें। / In \(9,18,27,\ldots\), (9n<200), so the greatest (n=22) and the term is (198). For multiples, take the largest integer below the limit.
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B. (448)
इस AP में (a=105), (d=7), इसलिए \(a_{50}=105+49\times7=448\)। गुणजों की AP में पहला सही गुणज चुनें। / In this AP, (a=105), (d=7), so \(a_{50}=105+49\times7=448\). In an AP of multiples, choose the first correct multiple.
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If the (3)rd term of an AP is (8) and the (8)th term is (3), what is the (11)th term?
A. (0)
\(d=\frac{3-8}{8-3}=-1\), इसलिए (a_{11}=3+3(-1)=0)। ज्ञात पास वाले पद से आगे बढ़ना सरल है। / \(d=\frac{3-8}{8-3}=-1\), so (a_{11}=3+3(-1)=0). Moving from the nearer known term is simple.
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If the (p)th term of an AP is (q) and the (q)th term is (p), what is its ((p+q))th term?
A. (0)
संबंधों को घटाने पर (d=-1) और आगे रखने पर \(a_{p+q}=0\) मिलता है। प्रतीकात्मक AP में भी (a_n=a+(n-1)d) ही लगाएं। / Subtracting the relations gives (d=-1), and substitution gives \(a_{p+q}=0\). Even in symbolic APs, use (a_n=a+(n-1)d).
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In the AP \(8,14,20,\ldots\), \(a_n=74\). Find the value of (n).
B. (12)
(74=8+(n-1)6) से (66=6(n-1)), इसलिए (n=12)। उत्तर में पद संख्या हमेशा पूर्णांक होनी चाहिए। / From (74=8+(n-1)6), (66=6(n-1)), so (n=12). The term number must always be an integer.
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In an AP, (a=4) and \(d=\frac{3}{2}\). What is \(a_{13}\)?
C. (22)
\(a_{13}=4+12\cdot\frac{3}{2}=22\)। भिन्न सार्व अंतर में गुणन को पहले सरल करें। / \(a_{13}=4+12\cdot\frac{3}{2}=22\). In fractional common differences, simplify multiplication first.
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If \(a_7=29\) and \(a_{14}=64\), what is the value of \(a_{21}\)?
D. (99)
\(d=\frac{64-29}{14-7}=5\), अतः \(a_{21}=64+7\times5=99\)। समान स्थान अंतर (7) का पद अंतर (35) होगा। / \(d=\frac{64-29}{14-7}=5\), hence \(a_{21}=64+7\times5=99\). Equal position gap (7) gives term gap (35).
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If the (10)th term of the AP \(x, x+4, x+8,\ldots\) is (50), what is the value of (x)?
C. (14)
\(50=x+9\times4\), इसलिए (x=14)। चर वाले पहले पद को (a) मानकर सूत्र लगाएं। / \(50=x+9\times4\), so (x=14). Treat the variable first term as (a) and apply the formula.
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B. (80)
(a=12), (d=4), अतः \(a_{18}=12+17\times4=80\)। सूची के पैटर्न से (d) पहले निकालें। / (a=12), (d=4), so \(a_{18}=12+17\times4=80\). First find (d) from the pattern.
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A. (240)
यहां (a=500), (d=-20), इसलिए (a_{14}=500+13(-20)=240)। घटते शब्द-प्रश्न में (d) ऋणात्मक लें। / Here (a=500), (d=-20), so (a_{14}=500+13(-20)=240). In decreasing word problems, take (d) as negative.
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B. (68.5) सेमी(68.5) cm
\(a_{30}=25+29\times1.5=68.5\) सेमी। पहले दिन को \(a_1\) मानें, दिन (0) नहीं। / \(a_{30}=25+29\times1.5=68.5\) cm. Treat the first day as \(a_1\), not day (0).
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B. (56)
\(a_{20}=18+19\times2=56\)। पंक्ति संख्या (20) होने पर अंतर (19) बार जुड़ता है। / \(a_{20}=18+19\times2=56\). For row number (20), the difference is added (19) times.
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B. (₹900)
(a=200), (d=50), अतः \(a_{15}=200+14\times50=900\)। हर सप्ताह की जमा राशि पद है, कुल राशि नहीं। / (a=200), (d=50), so \(a_{15}=200+14\times50=900\). The weekly deposit is the term, not the total amount.
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B. (43) सेमी(43) cm
यह AP है जिसमें (a=10), (d=3)। \(a_{12}=10+11\times3=43\) सेमी। शब्द-प्रश्न में पहले पद और अंतर पहचानें। / This is an AP with (a=10), (d=3). \(a_{12}=10+11\times3=43\) cm. In word problems, identify the first term and difference.
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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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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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