41 results found for "7m 3n" in Class 10.
C. (1240)
वांछित योग \(S_{25}-S_{15}=1975-735=1240\) है। परीक्षा में बीच के पदों के लिए कुल योगों का अंतर लें। / The required sum is \(S_{25}-S_{15}=1975-735=1240\). Exam tip: use the difference of cumulative sums for middle terms.
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If \(S_n=9n-3n^2\), what is the maximum value of the sum of initial terms?
A. (6)
\(S_1=6\), \(S_2=6\) और आगे योग घटता है इसलिए अधिकतम (6) है। परीक्षा में छोटे पूर्णांक मान जांचें। / \(S_1=6\), \(S_2=6\), and the sum decreases afterward, so the maximum is (6). Exam tip: check small integer values.
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If \(S_n=3n^2-2n\) for an arithmetic progression, what is its common difference?
B. (6)
\(a_1=S_1=1\) और \(a_2=S_2-S_1=7\) इसलिए (d=6)। परीक्षा में \(S_1\) और \(S_2-S_1\) से शुरू करें। / \(a_1=S_1=1\) and \(a_2=S_2-S_1=7\), so (d=6). Exam tip: start with \(S_1\) and \(S_2-S_1\).
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C. (199)
\(a_{25}=S_{25}-S_{24}=199\) है। परीक्षा में किसी पद के लिए लगातार दो योग घटाएं। / \(a_{25}=S_{25}-S_{24}=199\). Exam tip: subtract two consecutive sums to get a term.
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For an arithmetic progression \(S_n=5n^2-3n\). What is the (11)th term?
B. (102)
\(a_{11}=S_{11}-S_{10}\) इसलिए मान (572-470=102) है। परीक्षा में किसी पद के लिए \(S_n-S_{n-1}\) प्रयोग करें। / \(a_{11}=S_{11}-S_{10}\), so the value is (572-470=102). Exam tip: use \(S_n-S_{n-1}\) to find a particular term.
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A. (705)
दिए गए सूत्र में (n=15) रखने पर (S_{15}=3(15)2+2(15)=705)। परीक्षा में दिए गए \(S_n\) में सीधे (n) रखें। / Substituting (n=15) gives (S_{15}=3(15)2+2(15)=705). Exam tip: when \(S_n\) is given directly, substitute (n) first.
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If \(S_n=4n^2+3n\), find the sum from the (61)st term to the (90)th term.
A. (18090)
आवश्यक योग \(S_{90}-S_{60}=18090\) है। दिए गए \(S_n\) में सीमाओं के अनुसार सीधे घटाव करें। / The required sum is \(S_{90}-S_{60}=18090\). With given \(S_n\), subtract directly according to the limits.
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If \(S_n=3n^2+2n\), find the sum from the (51)st term to the (80)th term.
B. (11760)
आवश्यक योग \(S_{80}-S_{50}=11760\) है। \(S_n\) दिया हो तो सीमा-योग सीधे घटाव से निकालें। / The required sum is \(S_{80}-S_{50}=11760\). When \(S_n\) is given, find a range sum directly by subtraction.
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If the sum of an AP is \(S_n=8n^2-3n\), find the sum from the (51)st term to the (70)th term.
D. (19140)
आवश्यक योग \(S_{70}-S_{50}=19140\) है। \(S_n\) दिए होने पर सीमा-योग सीधे घटाव से निकालें। / The required sum is \(S_{70}-S_{50}=19140\). When \(S_n\) is given, find a range sum directly by subtraction.
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If the sum of an AP is \(S_n=3n^2+5n\), find the (22)nd term.
A. (134)
(22)वाँ पद \(S_{22}-S_{21}=134\) है। किसी एक पद को पाने के लिए \(S_n-S_{n-1}\) लगाएँ। / The (22)nd term is \(S_{22}-S_{21}=134\). To get a single term, use \(S_n-S_{n-1}\).
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If \(S_n=3n^2+4n\), find the sum from the (21)st term to the (30)th term.
C. (1540)
आवश्यक योग \(S_{30}-S_{20}=1540\) है। सीमा (21) से शुरू हो तो (20) तक का योग घटाएँ। / The required sum is \(S_{30}-S_{20}=1540\). If the range starts at (21), subtract the sum up to (20).
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D. (1089)
योग \(S_{20}-S_{11}=1089\) होगा। (12)वें से शुरू होने पर (11) पदों तक का योग घटाना होता है। / The sum is \(S_{20}-S_{11}=1089\). When starting from the (12)th term, subtract the sum up to (11) terms.
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If \(S_n=3n^2+2n\), what is the sum of the first (12) terms?
B. (456)
(n=12) रखने पर (S_{12}=3(12)2+2(12)=456)। दिए गए योग सूत्र में सीधे (n) का मान रखें। / Putting (n=12), (S_{12}=3(12)2+2(12)=456). Substitute the given value of (n) directly in the sum formula.
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D. (568)
पहला पद (13) और सोलहवाँ पद (58) है, इसलिए \(S_{16}=568\)। \(a_n\) से (a) और (l) निकालकर योग लगाएँ। / The first term is (13) and the sixteenth term is (58), so \(S_{16}=568\). Use \(a_n\) to find (a) and (l), then apply the sum formula.
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If \(S_n=3n^2+2n\), what will be the sum of the first (8) terms?
B. (208)
(S_8=3(8)2+2(8)=208)। \(S_n\) सीधे दिया हो तो (n) का मान रख दें। / (S_8=3(8)2+2(8)=208). If \(S_n\) is directly given, substitute the value of (n).
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If \(a_n=3n+2\), find the sum of the first (5) terms.
B. (55)
पहले (5) पद (5,8,11,14,17) हैं, जिनका योग (55) है। \(a_n\) दिया हो तो पद लिखकर जाँचें। / The first (5) terms are (5,8,11,14,17), whose sum is (55). If \(a_n\) is given, list the terms to check.
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If the sum of the first (n) terms is (S_n=\frac{n}{2}(3n+1)), what is \(S_{10}\)?
B. (155)
सूत्र में (n=10) रखें। (S_{10}=\frac{10}{2}(31)=155)। / Put (n=10) in the formula. (S_{10}=\frac{10}{2}(31)=155).
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If \(a_{3n+2}=148\), \(a_{n+2}=52\), and (d=12), what is (n)?
B. (4)
\(a_{3n+2}-a_{n+2}=2nd=96\)। (24n=96) से (n=4)। / \(a_{3n+2}-a_{n+2}=2nd=96\). From (24n=96), (n=4).
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If \(a_n=12n+q\) and \(a_{8n}-a_{3n}=780\), what is the value of (n)?
C. (13)
(a_{8n}-a_{3n}=(96n+q)-(36n+q)=60n)। (60n=780) से (n=13)। / (a_{8n}-a_{3n}=(96n+q)-(36n+q)=60n). From (60n=780), (n=13).
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In an AP \(a_{8n}=810\), \(a_{3n}=210\), and (d=24). What is (n)?
B. (5)
\(a_{8n}-a_{3n}=5nd=600\)। (120n=600) से (n=5)। / \(a_{8n}-a_{3n}=5nd=600\). From (120n=600), (n=5).
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If \(a_n=10n+q\) and \(a_{7n}-a_{3n}=520\), what is the value of (n)?
C. (13)
(a_{7n}-a_{3n}=(70n+q)-(30n+q)=40n)। (40n=520) से (n=13)। / (a_{7n}-a_{3n}=(70n+q)-(30n+q)=40n). From (40n=520), (n=13).
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In an AP, \(a_{7n}=530\), \(a_{3n}=146\), and (d=16). What is (n)?
B. (6)
\(a_{7n}-a_{3n}=4nd=384\)। (64n=384) से (n=6)। / \(a_{7n}-a_{3n}=4nd=384\). From (64n=384), (n=6).
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If \(a_n=13n-6\), what is (k) for \(a_{5k}-a_{2k}=234\)?
B. (6)
(a_{5k}-a_{2k}=(65k-6)-(26k-6)=39k)। (39k=234) से (k=6)। / (a_{5k}-a_{2k}=(65k-6)-(26k-6)=39k). From (39k=234), (k=6).
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If \(a_1=11\), (d=8), and \(a_{3n+4}=275\), what is the value of (n)?
B. (10)
(275=11+(3n+3)8) से (264=24n+24), इसलिए (n=10)। सूचकांक से (1) घटाकर (d) का गुणक बनाएं। / From (275=11+(3n+3)8), (264=24n+24), so (n=10). Subtract (1) from the index to get the multiplier of (d).
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If \(a_1=8\), (d=9), and \(a_{3n+2}=197\), what is the value of (n)?
B. (7)
(197=8+(3n+1)9) से (189=27n+9) इसलिए \(n=\frac{20}{3}\) नहीं आता। सूचकांक ठीक पढ़ना जरूरी है। / (197=8+(3n+1)9), so (189=27n+9) and \(n=\frac{20}{3}\). Read the index carefully before selecting an integer option.
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If \(a_n=13n+c\) and \(a_6=101\), what is (r) when \(a_{4r}=465\)?
B. (9)
(101=78+c) से (c=23)। (465=52r+23) से \(r=\frac{442}{52}=8.5\)। / From (101=78+c), (c=23). (465=52r+23), so \(r=\frac{442}{52}=8.5\).
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If \(a_n=3n+4\) and \(a_m=91\), what is \(a_{m+12}\)?
C. (127)
(3m+4=91) से (m=29)। \(a_{41}=3\times41+4=127\)। / From (3m+4=91), (m=29). \(a_{41}=3\times41+4=127\).
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If \(a_n=4n+q\) and \(a_{3n}-a_n=96\), what is the value of (n)?
C. (12)
(a_{3n}-a_n=(12n+q)-(4n+q)=8n)। (8n=96) से (n=12)। / (a_{3n}-a_n=(12n+q)-(4n+q)=8n). From (8n=96), (n=12).
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In an AP, \(a_{3n}=81\), \(a_n=21\), and (d=5). What is (n)?
B. (6)
\(a_{3n}-a_n=2nd=60\)। (10n=60) से (n=6)। / \(a_{3n}-a_n=2nd=60\). From (10n=60), (n=6).
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If \(a_n=3n+p\) and \(a_{12}=50\), what is \(a_{27}\)?
B. (95)
(50=36+p) से (p=14) और \(a_{27}=81+14=95\)। पहले स्थिरांक निकालें फिर मांगा पद रखें। / From (50=36+p), (p=14) and \(a_{27}=81+14=95\). First find the constant, then substitute the required term number.
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If \(a_n=10-3n\), which term will be (-44)?
C. (18)वां(18)th
(-44=10-3n) से (3n=54) इसलिए (n=18)। प्रत्यक्ष सूत्र में समीकरण को सीधे हल करें। / From (-44=10-3n), (3n=54) so (n=18). Solve the direct formula equation directly.
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If \(a_n=13n-6\), what is the value of \(a_{18}-a_{11}\)?
C. (91)
स्थान अंतर (18-11=7) और (d=13) है इसलिए अंतर \(7\times13=91\)। दोनों पद अलग से निकालना जरूरी नहीं। / The position gap is (18-11=7) and (d=13) so the difference is \(7\times13=91\). You do not need to find both terms separately.
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If \(a_n=3n-2\), what are the first term and common difference?
D. (a=1,d=3)
(n=1) रखने पर \(a_1=1\) और (n) का गुणांक (3) ही (d) है। प्रत्यक्ष सूत्र से दोनों मान तुरंत मिलते हैं। / Putting (n=1), \(a_1=1\) and the coefficient of (n) is (d=3). The direct formula gives both values quickly.
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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 \(a_n=30-3n\), what will be the (6)th term?
D. (12)
\(a_6=30-3\times6=12\)। घटते हुए सामान्य पद में घटाव सावधानी से करें। / \(a_6=30-3\times6=12\). Be careful with subtraction in a decreasing general term.
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Is the sequence \(a_n=n^2-3n+2\) an AP?
C. नहीं, \(a_{n+1}-a_n=2n-2\) स्थिर नहीं हैNo, \(a_{n+1}-a_n=2n-2\) is not constant
यहां अंतर (n) पर निर्भर करता है, इसलिए स्थिर नहीं है। परीक्षा में द्विघात \(a_n\) सामान्यतः समांतर श्रेणी नहीं देता। / Here the difference depends on (n), so it is not constant. In exams, a quadratic \(a_n\) usually does not form an AP.
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If (x-2+2(m-4)x+\(m^2-7m+14\)=0) has real roots, which condition on (m) is correct?
A. \(m\leq2\)
पिछले विविक्तकर से (D=4(2-m)) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(m\leq2\)। / From the discriminant, (D=4(2-m)). For real roots \(D\geq0\), so \(m\leq2\).
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If (x-2+2(m-4)x+\(m^2-7m+14\)=0) has equal roots, what is the value of (m)?
A. (2)
यहाँ (D=4(m-4)2-4\(m^2-7m+14\)=4(2-m)) है। (D=0) से (m=2) मिलता है। / Here (D=4(m-4)2-4\(m^2-7m+14\)=4(2-m)). From (D=0), (m=2).
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If one root of (x-2-(m+7)x+7m=0) is (7), what is the other root?
A. (m)
मूलों का गुणनफल (7m) है और एक मूल (7) है। इसलिए दूसरा मूल \(\frac{7m}{7}=m\) है। / The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
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C. परिमेय मान्यता विरोधाभास देती हैThe rational assumption gives a contradiction
(p=3m) और (q=3n) से (3) दोनों का साझा गुणनखंड है। / (p=3m) and (q=3n) show that (3) is a common factor of both.
यह सरलतम रूप की शर्त के विपरीत है। / This contradicts the lowest-form condition.
इसलिए \(\sqrt{3}\) को परिमेय मानना गलत सिद्ध होता है। / Therefore assuming \(\sqrt{3}\) rational is proved false.
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If (a) and (b) are coprime but the proof gives (a=3m) and (b=3n), what conclusion follows?
A. मान्यता में विरोधाभास हैThere is a contradiction in the assumption
(a=3m) और (b=3n) से दोनों (3) से विभाज्य हैं। / (a=3m) and (b=3n) show that both are divisible by (3).
इससे (3) साझा गुणनखंड बनता है। / Thus (3) becomes a common factor.
यह सहअभाज्य होने की शुरुआत वाली शर्त से टकराता है। / This conflicts with the starting condition of coprimality.
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