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1430 questions tagged with ap.
What will be the (24)th term of the AP \(-15,-9,-3,\ldots\)?
B. (123)
यहां (a=-15) और (d=6) है इसलिए \(a_{24}=-15+23\times6=123\)। ऋणात्मक पहले पद को अलग से ध्यान में रखें। / Here (a=-15) and (d=6) so \(a_{24}=-15+23\times6=123\). Keep the negative first term in mind.
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Which term of the AP \(21,29,37,\ldots\) is (149)?
C. (17)वां(17)th
(149=21+(n-1)8) से (128=8(n-1)) इसलिए (n=17)। पद संख्या पूछी हो तो दिए पद को \(a_n\) मानें। / From (149=21+(n-1)8), (128=8(n-1)) so (n=17). When term number is asked, treat the given term as \(a_n\).
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The (9)th term of an AP is (47) and the common difference is (4). What is the first term?
D. (15)
\(47=a+8\times4\) से (a=15)। ज्ञात पद से पहले पद तक जाने के लिए (8d) घटाएं। / From \(47=a+8\times4\), (a=15). To move from the known term to the first term, subtract (8d).
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If (a=13), (d=6), and \(a_n=121\), what is the value of (n)?
C. (19)
(121=13+(n-1)6) से (108=6(n-1)) इसलिए (n=19)। (n-1) मिलने के बाद (1) जरूर जोड़ें। / From (121=13+(n-1)6), (108=6(n-1)) so (n=19). After finding (n-1), add (1).
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Find the (17)th term of the AP \(38,33,28,\ldots\).
A. (-42)
यहां (a=38) और (d=-5) है इसलिए (a_{17}=38+16(-5)=-42)। घटती श्रेणी में (d) को ऋणात्मक लें। / Here (a=38) and (d=-5) so (a_{17}=38+16(-5)=-42). In a decreasing AP take (d) as negative.
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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, \(a_{10}=31\) and \(a_{18}=55\). What is \(a_6\)?
B. (19)
\(d=\frac{55-31}{8}=3\), इसलिए \(a_6=31-4\times3=19\)। ज्ञात पद से पीछे आने पर (d) घटाएं। / \(d=\frac{55-31}{8}=3\), so \(a_6=31-4\times3=19\). When moving backward from a known term, subtract (d).
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If \(a_n=50-4n\), which term of the AP will be (-10)?
C. (15)वां(15)th
(-10=50-4n) से (4n=60), अतः (n=15)। प्रत्यक्ष सूत्र में समीकरण को सीधे हल करें। / From (-10=50-4n), (4n=60), so (n=15). Solve the direct formula equation directly.
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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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The (n)th term of the AP \(27,24,21,\ldots\) is (-30). What is (n)?
C. (20)
(-30=27+(n-1)(-3)) से (57=3(n-1)), इसलिए (n=20)। ऋणात्मक लक्ष्य पद में चिन्ह सावधानी से बदलें। / From (-30=27+(n-1)(-3)), (57=3(n-1)), so (n=20). Handle signs carefully with a negative target term.
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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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In an AP, \(a_{15}=70\) and \(a_{25}=120\). What is \(a_1\)?
A. (0)
\(d=\frac{120-70}{25-15}=5\), इसलिए \(a_1=70-14\times5=0\)। पहले (d) निकालकर पीछे जाएं। / \(d=\frac{120-70}{25-15}=5\), so \(a_1=70-14\times5=0\). First find (d), then move backward.
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If the (n)th term of an AP is \(a_n=9n+2\), what is the value of \(a_{11}-a_5\)?
B. (54)
(a_{11}-a_5=(101)-(47)=54), या \(6d=6\times9=54\)। पदों के अंतर में स्थानों का अंतर उपयोग करें। / (a_{11}-a_5=(101)-(47)=54), or \(6d=6\times9=54\). Use the difference of positions for term differences.
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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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