100 results found for "ap term-position nth-term class10" in Class 10.
Which term of the AP \(9,17,25,\ldots\) is (201)?
C. (25)वां(25)th
(201=9+(n-1)8) से (192=8(n-1)) और (n=25)। पद संख्या पूर्णांक आए तो उत्तर सही दिशा में है। / From (201=9+(n-1)8), (192=8(n-1)) and (n=25). If the term number is an integer the answer is on the right track.
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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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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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If the (5)th term of an AP is (27) and the (14)th term is (90), what is the (20)th term?
C. (132)
\(d=\frac{90-27}{14-5}=7\) और \(a_{20}=90+6\times7=132\)। पहले (d) निकालें फिर आगे बढ़ें। / \(d=\frac{90-27}{14-5}=7\) and \(a_{20}=90+6\times7=132\). First find (d) then move forward.
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If the (4)th term of an AP is (15) and the (12)th term is (55), what is the (16)th term?
C. (75)
\(d=\frac{55-15}{12-4}=5\) और \(a_{16}=55+4\times5=75\)। पहले (d) फिर वांछित पद निकालें। / \(d=\frac{55-15}{12-4}=5\) and \(a_{16}=55+4\times5=75\). First find (d), then the required term.
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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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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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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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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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The (n)th-term formula of the AP \(23,31,39,\ldots\) is \(a_n=8n+15\). What is the (36)th term?
C. (303)
\(a_{36}=8\times36+15=303\)। बने हुए सूत्र में (n=36) रखकर सीधे उत्तर पाएं। / \(a_{36}=8\times36+15=303\). Put (n=36) in the formed formula to get the answer directly.
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If the (r)th term of an AP is (3r-2) and (d=4), what is the ((r+6))th term?
C. (3r+22)
((r+6))वां पद (r)वें पद से (6d) आगे है इसलिए (3r-2+24=3r+22)। प्रतीकात्मक प्रश्न में स्थान अंतर देखें। / The ((r+6))th term is (6d) ahead of the (r)th term so (3r-2+24=3r+22). In symbolic questions look at the position gap.
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The (n)th-term formula of the AP \(19,26,33,\ldots\) is \(a_n=7n+12\). What is the (41)st term?
D. (299)
\(a_{41}=7\times41+12=299\)। बने हुए सूत्र में केवल (n=41) रखें। / \(a_{41}=7\times41+12=299\). Put only (n=41) in the formed formula.
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If the (m)th term of an AP is (2m+3) and (d=2), what is the ((m+5))th term?
D. (2m+13)
((m+5))वां पद (m)वें पद से (5d) आगे है इसलिए (2m+3+10=2m+13)। प्रतीकात्मक पदों में भी स्थान अंतर देखें। / The ((m+5))th term is (5d) ahead of the (m)th term, so (2m+3+10=2m+13). In symbolic terms, look at the position gap.
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The first term of an AP is (25) and the (18)th term is (93). What is the common difference?
C. (4)
(93=25+17d) से (68=17d) इसलिए (d=4)। (18)वें पद के लिए गुणक (17) होता है। / From (93=25+17d), (68=17d) so (d=4). For the (18)th term, the multiplier is (17).
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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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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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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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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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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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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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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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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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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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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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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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If the first term of an AP is (5) and common difference is (3), what is the (8)th term?
C. (26)
सूत्र (a_n=a+(n-1)d) लगाएं तो \(5+7\times3=26\)। परीक्षा में (n-1) लेना न भूलें। / Using (a_n=a+(n-1)d), \(5+7\times3=26\). In exams, do not forget (n-1).
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If the (19)th term of the AP \(z,z+5,z+10,\ldots\) is (112), what is (z)?
C. (22)
\(112=z+18\times5\) से (z=22)। चर वाले पहले पद को (a) मानकर सूत्र लगाएं। / From \(112=z+18\times5\), (z=22). Treat the variable first term as (a) and apply the formula.
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In the AP \(7,12,17,\ldots\), what is the last term less than (180)?
C. (177)
पद (7+5(n-1)) हैं। (180) से कम अंतिम पद (177) है क्योंकि अगला पद (182) होगा। / The terms are (7+5(n-1)). The last term less than (180) is (177) because the next term will be (182).
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Which is the first negative term of the AP \(105,98,91,\ldots\)?
C. (17)वां(17)th
(a_n=105+(n-1)(-7)=112-7n)। \(a_n<0\) से (n>16) इसलिए पहला ऋणात्मक पद (17)वां है। / (a_n=105+(n-1)(-7)=112-7n). From \(a_n<0\), (n>16) so the first negative term is the (17)th.
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Which term of the AP \(26,35,44,\ldots\) is (206)?
C. (21)वां(21)st
(206=26+(n-1)9) से (180=9(n-1)) इसलिए (n=21)। पद और पहले पद का अंतर (d) से भाग दें। / From (206=26+(n-1)9), (180=9(n-1)) so (n=21). Divide the difference between the term and first term by (d).
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If the (12)th term of the AP \(y,y+6,y+12,\ldots\) is (89), what is the value of (y)?
D. (23)
\(89=y+11\times6\) से (y=23)। चर वाले पहले पद को सीधे (a) मानें। / From \(89=y+11\times6\), (y=23). Treat the variable first term directly as (a).
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In the AP \(2,6,10,\ldots\), what is the last term less than (75)?
A. (74)
इस श्रेणी के पद (2+4(n-1)) हैं। (75) से कम अंतिम पद (74) है। / The terms of this sequence are (2+4(n-1)). The last term less than (75) is (74).
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Which is the first negative term of the AP \(72,66,60,\ldots\)?
B. (14)वां(14)th
(a_n=72+(n-1)(-6)=78-6n)। \(a_n<0\) से (n>13) इसलिए पहला ऋणात्मक पद (14)वां है। / (a_n=72+(n-1)(-6)=78-6n). From \(a_n<0\), (n>13) so the first negative term is the (14)th.
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Which term of the AP \(14,23,32,\ldots\) is (185)?
C. (20)वां(20)th
(185=14+(n-1)9) से (171=9(n-1)) और (n=20)। अंतर को (d) से भाग देकर पद संख्या पाएं। / From (185=14+(n-1)9), (171=9(n-1)) and (n=20). Divide the difference by (d) to get the term number.
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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=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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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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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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The general term of an AP is \(a_n=7n+2\). Find the (11)th term.
C. (79)
\(a_{11}=7\times11+2=79\)। सामान्य पद में सही पद संख्या रखना ही मुख्य कदम है। / \(a_{11}=7\times11+2=79\). The main step is substituting the correct term number in the general term.
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The AP of multiples of (12) greater than (400) is \(408,420,432,\ldots\). What is its (18)th term?
A. (612)
यहां (a=408) और (d=12) है इसलिए \(a_{18}=408+17\times12=612\)। सीमा के बाद पहला सही गुणज चुनें। / Here (a=408) and (d=12) so \(a_{18}=408+17\times12=612\). Choose the first correct multiple after the limit.
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In the AP \(13,21,29,\ldots\), what is the greatest term less than (250)?
C. (245)
पद (13+8(n-1)) के रूप में हैं और (250) से कम सबसे बड़ा ऐसा पद (245) है। सीमा वाले प्रश्न में अगले पद से भी जांच करें। / The terms are of the form (13+8(n-1)) and the greatest such term less than (250) is (245). In limit questions check the next term too.
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If the first term of an AP is (22) and \(a_{32}=177\), what is the common difference?
C. (5)
(177=22+31d) से (155=31d) और (d=5)। (32)वें पद में (31d) जुड़ता है। / From (177=22+31d), (155=31d) and (d=5). In the (32)nd term (31d) is added.
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What will be the (20)th term of the AP \(-45,-36,-27,\ldots\)?
C. (126)
(a=-45) और (d=9) हैं इसलिए \(a_{20}=-45+19\times9=126\)। ऋणात्मक शुरूआत वाली बढ़ती AP में अंतिम जोड़ सावधानी से करें। / (a=-45) and (d=9) so \(a_{20}=-45+19\times9=126\). In an increasing AP starting negative add carefully at the end.
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What is the (28)th term of the AP \(\frac{2}{3},\frac{4}{3},2,\ldots\)?
B. \(\frac{56}{3}\)
यहां \(a=\frac{2}{3}\) और \(d=\frac{2}{3}\) है इसलिए \(a_{28}=\frac{2}{3}+27\cdot\frac{2}{3}=\frac{56}{3}\)। भिन्नों में गुणन को पहले सरल करें। / Here \(a=\frac{2}{3}\) and \(d=\frac{2}{3}\) so \(a_{28}=\frac{2}{3}+27\cdot\frac{2}{3}=\frac{56}{3}\). Simplify multiplication first in fractions.
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Find the (16)th term of the AP \(2.4,4.0,5.6,\ldots\).
C. (26.4)
यहां (a=2.4) और (d=1.6) है इसलिए \(a_{16}=2.4+15\times1.6=26.4\)। दशमलव गुणा में स्थान मूल्य ध्यान रखें। / Here (a=2.4) and (d=1.6) so \(a_{16}=2.4+15\times1.6=26.4\). Keep place value in mind while multiplying decimals.
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The first term of an AP is (31) and \(a_{19}=139\). What is the common difference?
C. (6)
(139=31+18d) से (108=18d) इसलिए (d=6)। (19)वें पद के लिए (18d) जुड़ता है। / From (139=31+18d), (108=18d) so (d=6). For the (19)th term (18d) is added.
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What will be the (13)th term of the AP \(120,109,98,\ldots\)?
C. (-12)
(d=-11) है इसलिए (a_{13}=120+12(-11)=-12)। बड़े ऋणात्मक अंतर में गुणा पहले करें। / (d=-11) so (a_{13}=120+12(-11)=-12). With a large negative difference multiply first.
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What will be the (21)st term of the AP \(-28,-20,-12,\ldots\)?
D. (132)
यहां (a=-28) और (d=8) है इसलिए \(a_{21}=-28+20\times8=132\)। ऋणात्मक पहले पद को जोड़ते समय सावधानी रखें। / Here (a=-28) and (d=8) so \(a_{21}=-28+20\times8=132\). Be careful while adding a negative first term.
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Find the (18)th term of the AP \(84,78,72,\ldots\).
B. (-18)
यहां (d=-6) है इसलिए (a_{18}=84+17(-6)=-18)। घटती AP में सार्व अंतर को ऋणात्मक लें। / Here (d=-6) so (a_{18}=84+17(-6)=-18). In a decreasing AP take the common difference as negative.
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What is the (23)rd term of the AP \(6,14,22,\ldots\)?
A. (182)
यहां (a=6) और (d=8) है इसलिए \(a_{23}=6+22\times8=182\)। परीक्षा में (n) की जगह (n-1) लगाना याद रखें। / Here (a=6) and (d=8) so \(a_{23}=6+22\times8=182\). Remember to use (n-1) instead of (n) in exams.
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The AP of multiples of (8) greater than (300) is \(304,312,320,\ldots\). What is its (25)th term?
A. (496)
यहां (a=304) और (d=8) है इसलिए \(a_{25}=304+24\times8=496\)। सीमा के बाद पहला सही गुणज चुनें। / Here (a=304) and (d=8) so \(a_{25}=304+24\times8=496\). Choose the first correct multiple after the limit.
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In the AP \(5,12,19,\ldots\), what is the greatest term less than (150)?
A. (145)
पद (5+7(n-1)) के रूप में हैं। (150) से कम सबसे बड़ा ऐसा पद (145) है। / The terms are of the form (5+7(n-1)). The greatest such term less than (150) is (145).
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If the first term of an AP is (16) and \(a_{26}=116\), what is (d)?
A. (4)
(116=16+25d) से (100=25d) इसलिए (d=4)। (26)वें पद में (d) (25) बार जुड़ता है। / From (116=16+25d), (100=25d) so (d=4). In the (26)th term, (d) is added (25) times.
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What will be the (22)nd term of the AP \(-32,-25,-18,\ldots\)?
B. (115)
(a=-32) और (d=7) हैं इसलिए \(a_{22}=-32+21\times7=115\)। ऋणात्मक आरंभ में जोड़ते समय सावधान रहें। / (a=-32) and (d=7) so \(a_{22}=-32+21\times7=115\). Be careful while adding from a negative start.
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What is the (31)st term of the AP \(\frac{3}{4},\frac{5}{4},\frac{7}{4},\ldots\)?
B. \(\frac{63}{4}\)
यहां \(a=\frac{3}{4}\) और \(d=\frac{1}{2}\) है इसलिए \(a_{31}=\frac{3}{4}+30\cdot\frac{1}{2}=\frac{63}{4}\)। भिन्नों में समान हर बनाएं। / Here \(a=\frac{3}{4}\) and \(d=\frac{1}{2}\) so \(a_{31}=\frac{3}{4}+30\cdot\frac{1}{2}=\frac{63}{4}\). Use a common denominator for fractions.
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Find the (19)th term of the AP \(1.2,2.7,4.2,\ldots\).
D. (28.2)
यहां (a=1.2) और (d=1.5) है इसलिए \(a_{19}=1.2+18\times1.5=28.2\)। दशमलव में गुणा सावधानी से करें। / Here (a=1.2) and (d=1.5) so \(a_{19}=1.2+18\times1.5=28.2\). Multiply decimals carefully.
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What is the (12)th term of the AP \(96,87,78,\ldots\)?
A. (-3)
(d=-9) है इसलिए (a_{12}=96+11(-9)=-3)। बड़े घटते अंतर में चिन्ह गलती से न बदलें। / (d=-9) so (a_{12}=96+11(-9)=-3). Do not accidentally change the sign in a large decreasing difference.
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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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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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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 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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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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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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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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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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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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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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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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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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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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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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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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