Level 53 • 34/50 questions • 25 seconds per question.
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If a geometric progression has \(a_2=12\) and (r=4), what will be \(a_5\)?
D. (768)
\(a_5=a_2r^3=12\cdot4^3=768\)। दिए पद से आगे जाने में अनुपात की सही घात लगाएं। / \(a_5=a_2r^3=12\cdot4^3=768\). Use the correct power of the ratio when moving forward from a given term.
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What is the sixth term of the geometric progression \(5,-15,45,-135,\ldots\)?
C. (-1215)
सार्व अनुपात (-3) है और पद (5,-15,45,-135,405,-1215) हैं। ऋणात्मक अनुपात में चिह्न बदलते हैं। / The common ratio is (-3), and the terms are (5,-15,45,-135,405,-1215). With a negative ratio, signs alternate.
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If (4,x,100) are consecutive terms of a positive geometric progression, what is the value of (x)?
A. (20)
\(x^2=4\cdot100=400\), इसलिए धनात्मक (x=20)। मध्य पद ज्यामितीय माध्य होता है। / \(x^2=4\cdot100=400\), so positive (x=20). The middle term is the geometric mean.
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In a geometric progression, \(a_3=24\) and \(a_6=648\). What is the value of \(a_4\)?
B. (72)
\(a_6=a_3r^3\), इसलिए \(648=24r^3\) और (r=3), अतः \(a_4=72\)। पहले अनुपात निकालें फिर अगला पद बनाएं। / \(a_6=a_3r^3\), so \(648=24r^3\) and (r=3), hence \(a_4=72\). First find the ratio and then form the next term.
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If \(a_n=2\cdot3^{n-1}\), what will be the value of \(a_4+a_6\)?
C. (540)
\(a_4=54\) और \(a_6=486\), इसलिए योग (540) है। दोनों पद अलग निकालकर जोड़ें। / \(a_4=54\) and \(a_6=486\), so the sum is (540). Find both terms separately and add them.
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In the geometric progression \(160,80,40,20,\ldots\), which term is \(\frac{5}{2}\)?
B. (7)वाँ(7)th
पद \(160,80,40,20,10,5,\frac{5}{2}\) हैं, इसलिए यह सातवाँ पद है। घटती श्रेढ़ी में अनुपात से बार-बार गुणा करें। / The terms are \(160,80,40,20,10,5,\frac{5}{2}\), so it is the seventh term. In a decreasing progression, multiply repeatedly by the ratio.
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If \(a_2=10\) and \(a_6=160\), and the ratio is positive, what will be \(a_1\)?
D. (5)
\(a_6=a_2r^4\), इसलिए \(160=10r^4\) और (r=2), अतः \(a_1=5\)। पीछे जाने के लिए अनुपात से भाग दें। / \(a_6=a_2r^4\), so \(160=10r^4\) and (r=2), hence \(a_1=5\). Divide by the ratio to move backward.
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If (x-2,x+4,2x+8) are consecutive positive terms of a geometric progression, what is (x)?
A. (8)
मध्य पद के लिए ((x+4)2=(x-2)(2x+8)), इससे (x=8) मिलता है। तीन पदों में मध्य पद का वर्ग दोनों किनारों के गुणनफल के बराबर होता है। / For the middle term, ((x+4)2=(x-2)(2x+8)), which gives (x=8). In three GP terms, the square of the middle term equals the product of the extremes.
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In the geometric progression \(3,6,12,\ldots\), what is the greatest term less than (100)?
C. (96)
पद (3,6,12,24,48,96,192) हैं, इसलिए (100) से छोटा सबसे बड़ा पद (96) है। सीमा वाले प्रश्न में पास के पद जांचें। / The terms are (3,6,12,24,48,96,192), so the greatest term below (100) is (96). In boundary questions, check nearby terms.
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B. (5)
\(a_2=2a_1\) और \(a_3=4a_1\), इसलिए \(6a_1=30\) और \(a_1=5\)। योग को पहले \(a_1\) और (r) से लिखें। / \(a_2=2a_1\) and \(a_3=4a_1\), so \(6a_1=30\) and \(a_1=5\). First write the sum using \(a_1\) and (r).
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If \(a_1=2\) and the sum of the first (5) terms is (62), and (r) is a positive integer, what is (r)?
A. (2)
(r=2) पर पद (2,4,8,16,32) हैं और योग (62) है। छोटे विकल्पों में पद बनाकर जांचना तेज होता है। / For (r=2), the terms are (2,4,8,16,32) and the sum is (62). For small options, forming the terms is quick.
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In a geometric progression, \(a_2=15\) and \(a_5=405\). For positive (r), what will \(a_1\) be?
C. (5)
\(a_5=a_2r^3\), इसलिए \(405=15r^3\) और (r=3), अतः \(a_1=5\)। पहले अनुपात निकालें फिर पीछे जाएं। / \(a_5=a_2r^3\), so \(405=15r^3\) and (r=3), hence \(a_1=5\). First find the ratio and then move backward.
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If (2,x,18,y) is a geometric progression and all terms are positive, what will (y) be?
C. (54)
\(x^2=2\cdot18=36\), इसलिए (x=6) और (r=3), अतः (y=54)। पहले मध्य पद से अनुपात निकालें। / \(x^2=2\cdot18=36\), so (x=6) and (r=3), hence (y=54). First use the middle term to find the ratio.
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If \(a_1=\frac{3}{2}\) and (r=4), what will be \(a_5\)?
B. (384)
\(a_5=\frac{3}{2}\cdot4^4=384\)। भिन्न पहले पद के साथ घात का गुणा ध्यान से करें। / \(a_5=\frac{3}{2}\cdot4^4=384\). Multiply the power carefully with the fractional first term.
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In the geometric progression \(1,3,9,27,\ldots\), what is the greatest term less than (1000)?
B. (729)
पद (1,3,9,27,81,243,729,2187) हैं, इसलिए (1000) से छोटा सबसे बड़ा पद (729) है। सीमा वाले प्रश्न में अगला पद भी जांचें। / The terms are (1,3,9,27,81,243,729,2187), so the greatest term below (1000) is (729). In boundary questions, check the next term too.
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If \(6,18,54,\ldots\) is a geometric progression, what is the value of \(a_5-a_3\)?
A. (432)
\(a_3=54\) और \(a_5=486\), इसलिए अंतर (432) है। दोनों पद अलग निकालकर घटाएं। / \(a_3=54\) and \(a_5=486\), so the difference is (432). Find both terms separately and subtract.
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In the geometric progression \(a,ar,ar^2,\ldots\), \(a_2=8\) and \(a_4=128\). What will be \(a_6\)?
C. (2048)
\(a_4=a_2r^2\), इसलिए \(r^2=16\) और धनात्मक क्रम में \(a_6=128\cdot16=2048\)। समान दूरी पर वही \(r^2\) फिर लगता है। / \(a_4=a_2r^2\), so \(r^2=16\), and for the positive sequence \(a_6=128\cdot16=2048\). The same \(r^2\) applies over the same gap.
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If \(a_n=81\left(\frac{1}{3}\right)^{n-1}\), what is \(a_2+a_5\)?
A. (28)
\(a_2=27\) और \(a_5=1\), इसलिए योग (28) है। भिन्न अनुपात में हर पद तेजी से घटता है। / \(a_2=27\) and \(a_5=1\), so the sum is (28). With a fractional ratio, terms decrease quickly.
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If \(a_1=4\) and (r=-2), what will be the sum of the first (5) terms?
D. (44)
पद (4,-8,16,-32,64) हैं और योग (44) है। ऋणात्मक अनुपात में चिह्नों को ध्यान से जोड़ें। / The terms are (4,-8,16,-32,64), and their sum is (44). Add signs carefully with a negative ratio.
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If a geometric progression has \(a_2=6\) and \(a_4=54\), what will be the value of \(a_1+a_5\)?
A. (164)
\(a_4=a_2r^2\), इसलिए (r=3), \(a_1=2\) और \(a_5=162\), अतः योग (164) है। पहले (r) निकालें। / \(a_4=a_2r^2\), so (r=3), \(a_1=2\), and \(a_5=162\), hence the sum is (164). Find (r) first.
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If \(a_n=5\cdot2^{n-1}\), what will (n) be for \(a_n=320\)?
C. (7)
\(5\cdot2^{n-1}=320\), इसलिए \(2^{n-1}=64=2^6\) और (n=7)। घातों की तुलना करें। / \(5\cdot2^{n-1}=320\), so \(2^{n-1}=64=2^6\) and (n=7). Compare powers.
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The consecutive terms of a geometric progression are (x,3x,27). What is the positive (x)?
C. (3)
((3x)2=x\cdot27), इसलिए \(9x^2=27x\) और धनात्मक (x=3)। शून्य को धनात्मक हल में नहीं लें। / ((3x)2=x\cdot27), so \(9x^2=27x\) and positive (x=3). Do not take zero as a positive solution.
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If \(a_1=6\) and (r=3), what is \(a_6-a_4\)?
C. (1296)
\(a_4=162\) और \(a_6=1458\), इसलिए अंतर (1296) है। पहले दोनों पद सही घात से निकालें। / \(a_4=162\) and \(a_6=1458\), so the difference is (1296). First find both terms using the correct powers.
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In the geometric progression \(2,10,50,\ldots\), what is the greatest term less than (6000)?
A. (1250)
पद (2,10,50,250,1250,6250) हैं, इसलिए (6000) से छोटा सबसे बड़ा पद (1250) है। अगला पद सीमा पार करता है। / The terms are (2,10,50,250,1250,6250), so the greatest term below (6000) is (1250). The next term crosses the limit.
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If \(a_3=12\) and \(a_6=96\), and the ratio is positive, what will \(a_8\) be?
C. (384)
\(a_6=a_3r^3\), इसलिए (r=2), अतः \(a_8=96\cdot2^2=384\)। पहले अनुपात निकालकर आगे बढ़ें। / \(a_6=a_3r^3\), so (r=2), hence \(a_8=96\cdot2^2=384\). First find the ratio and then move forward.
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If \(a_1=1\) and (r=4), what will be the product of the first (4) terms?
C. (4096)
पहले चार पद (1,4,16,64) हैं और गुणनफल (4096) है। गुणनफल में हर पद को सावधानी से गुणा करें। / The first four terms are (1,4,16,64), and their product is (4096). Multiply each term carefully in product questions.
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If \(a_n=7\cdot3^{n-1}\), what is the value of \(\frac{a_5}{a_3}\)?
C. (9)
\(\frac{a_5}{a_3}=r^2=3^2=9\)। समान श्रेढ़ी में पदों का अनुपात पद-संख्या के अंतर पर निर्भर करता है। / \(\frac{a_5}{a_3}=r^2=3^2=9\). In the same progression, the ratio of terms depends on the difference of positions.
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If \(a_2=20\) and \(a_4=80\), and (r) is positive, what will be \(a_6\)?
C. (320)
\(a_4=a_2r^2\), इसलिए \(r^2=4\), और \(a_6=a_4r^2=80\cdot4=320\)। समान दूरी पर वही गुणक लगता है। / \(a_4=a_2r^2\), so \(r^2=4\), and \(a_6=a_4r^2=80\cdot4=320\). The same multiplier applies over the same gap.
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For which (x) will (x,12,48) be consecutive terms of a geometric progression?
B. (3)
\(\frac{48}{12}=4\), इसलिए \(\frac{12}{x}=4\) और (x=3)। लगातार अनुपात बराबर रखें। / \(\frac{48}{12}=4\), so \(\frac{12}{x}=4\) and (x=3). Keep consecutive ratios equal.
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What is the sum of the first (5) terms of the geometric progression \(12,6,3,\frac{3}{2},\ldots\)?
B. \(\frac{93}{4}\)
पहले (5) पद \(12,6,3,\frac{3}{2},\frac{3}{4}\) हैं और योग \(\frac{93}{4}\) है। भिन्न पदों को समान हर से जोड़ें। / The first (5) terms are \(12,6,3,\frac{3}{2},\frac{3}{4}\), and their sum is \(\frac{93}{4}\). Add fractional terms using a common denominator.
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If \(a_1=11\) and (r=3), what will \(a_3+a_5\) be?
A. (990)
\(a_3=99\) और \(a_5=891\), इसलिए योग (990) है। दोनों पद अलग निकालें। / \(a_3=99\) and \(a_5=891\), so the sum is (990). Find both terms separately.
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If \(a_2=14\) and (r=5), what is the value of \(a_1+a_3\)?
B. \(\frac{364}{5}\)
\(a_1=\frac{14}{5}\) और \(a_3=70\), इसलिए योग \(\frac{364}{5}\) है। आगे और पीछे दोनों दिशा में अनुपात का उपयोग करें। / \(a_1=\frac{14}{5}\) and \(a_3=70\), so the sum is \(\frac{364}{5}\). Use the ratio both backward and forward.
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If (16,x,81) are consecutive terms of a positive geometric progression, what will (x) be?
C. (36)
\(x^2=16\cdot81=1296\), इसलिए धनात्मक (x=36)। मध्य पद का वर्ग बाहरी पदों के गुणनफल के बराबर होता है। / \(x^2=16\cdot81=1296\), so positive (x=36). The square of the middle term equals the product of the outer terms.
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If \(a_1=5\) and \(a_4=625\), and (r) is positive, what will \(a_5\) be?
D. (3125)
\(625=5r^3\), इसलिए \(r^3=125\) और (r=5), अतः \(a_5=3125\)। पहले अनुपात निकालें फिर अगला पद बनाएं। / \(625=5r^3\), so \(r^3=125\) and (r=5), hence \(a_5=3125\). First find the ratio and then form the next term.
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