Level 55 • 28/50 questions • 25 seconds per question.
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What is the tenth term of the geometric progression \(2,6,18,54,\ldots\)?
B. (39366)
यहाँ (a=2) और (r=3) है, इसलिए \(a_{10}=2\cdot3^9=39366\) है। परीक्षा में \(a_n=ar^{n-1}\) लगाएँ। / Here (a=2) and (r=3), so \(a_{10}=2\cdot3^9=39366\). In exams, use \(a_n=ar^{n-1}\).
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B. (186)
\(\frac{192}{24}=r^3=8\) से (r=2) और (a=6) है। इसलिए (S_5=6\(2^5-1\)=186) होगा। / From \(\frac{192}{24}=r^3=8\), (r=2) and (a=6). Therefore (S_5=6\(2^5-1\)=186).
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What is the ninth term of the geometric progression \(1024,512,256,128,\ldots\)?
B. (4)
यहाँ \(r=\frac{1}{2}\) है, इसलिए (a_9=1024\cdot\left\(\frac{1}{2}\right\)8=4) है। परीक्षा में घटती GP में भिन्न अनुपात सावधानी से लगाएँ। / Here \(r=\frac{1}{2}\), so (a_9=1024\cdot\left\(\frac{1}{2}\right\)8=4). In exams, apply fractional ratios carefully in decreasing GPs.
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If \(a_n=7\cdot2^{n-1}\), which term will be (1792)?
B. नौवाँ पद(9)th term
\(7\cdot2^{n-1}=1792\) से \(2^{n-1}=256=2^8\), इसलिए (n=9) है। परीक्षा में घातों को बराबर करके पद संख्या निकालें। / From \(7\cdot2^{n-1}=1792\), \(2^{n-1}=256=2^8\), so (n=9). In exams, equate powers to find the term number.
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B. (3)
\(324=4r^4\) से \(r^4=81=3^4\), इसलिए (r=3) है। परीक्षा में पाँचवें पद के लिए \(r^4\) याद रखें। / From \(324=4r^4\), \(r^4=81=3^4\), so (r=3). In exams, remember \(r^4\) for the fifth term.
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What is the general term of the geometric progression \(6,24,96,384,\ldots\)?
A. \(a_n=6\cdot4^{n-1}\)
पहला पद (6) और अनुपात (4) है, इसलिए \(a_n=6\cdot4^{n-1}\) है। परीक्षा में \(ar^{n-1}\) में (a) और (r) सही रखें। / The first term is (6) and the ratio is (4), so \(a_n=6\cdot4^{n-1}\). In exams, keep (a) and (r) correct in \(ar^{n-1}\).
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What is the sum of the first (7) terms of the geometric progression \(3,9,27,81,\ldots\)?
B. (3279)
(S_7=\frac{3\(3^7-1\)}{3-1}=3279) है। परीक्षा में (r>1) के लिए (\frac{a\(r^n-1\)}{r-1}) लगाएँ। / (S_7=\frac{3\(3^7-1\)}{3-1}=3279). In exams, use (\frac{a\(r^n-1\)}{r-1}) for (r>1).
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What is the sum of the first (8) terms of the geometric progression \(8,16,32,64,\ldots\)?
C. (2040)
(S_8=8\(2^8-1\)=2040) है। परीक्षा में अनुपात (2) हो तो \(2^n\) जल्दी निकालें। / (S_8=8\(2^8-1\)=2040). In exams, calculate \(2^n\) quickly when the ratio is (2).
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If \(a_5=625\) and (r=5), what is the first term (a)?
A. (1)
\(a_5=ar^4\), इसलिए \(625=a\cdot5^4\) और (a=1) है। परीक्षा में पाँचवें पद से \(r^4\) भाग दें। / \(a_5=ar^4\), so \(625=a\cdot5^4\) and (a=1). In exams, divide the fifth term by \(r^4\).
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What is the general term of the geometric progression \(243,81,27,9,\ldots\)?
B. (a_n=243\cdot\left\(\frac{1}{3}\right\)^{n-1})
पहला पद (243) और अनुपात \(\frac{1}{3}\) है, इसलिए सही नियम (243\cdot\left\(\frac{1}{3}\right\)^{n-1}) है। परीक्षा में घटती GP में भिन्न अनुपात लिखें। / The first term is (243) and the ratio is \(\frac{1}{3}\), so the correct rule is (243\cdot\left\(\frac{1}{3}\right\)^{n-1}). In exams, write the fractional ratio in a decreasing GP.
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In the geometric progression \(5,20,80,320,\ldots\), which term is (5120)?
B. छठा पद(6)th term
\(5\cdot4^{n-1}=5120\) से \(4^{n-1}=1024=4^5\), इसलिए (n=6) है। परीक्षा में घातों को बराबर करके (n) निकालें। / From \(5\cdot4^{n-1}=5120\), \(4^{n-1}=1024=4^5\), so (n=6). In exams, equate powers to find (n).
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A. सही क्योंकि \(S_5=363\)True because \(S_5=363\)
(S_5=\frac{3\(3^5-1\)}{3-1}=363) है, इसलिए कथन सही है। परीक्षा में कथन की जाँच सूत्र से करें। / (S_5=\frac{3\(3^5-1\)}{3-1}=363), so the statement is true. In exams, verify the statement using the formula.
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What is the sum of the first (6) terms of the geometric progression \(1,4,16,64,\ldots\)?
A. (1365)
\(S_6=\frac{4^6-1}{4-1}=1365\) है। परीक्षा में (r>1) के लिए \(\frac{r^n-1}{r-1}\) रूप उपयोगी है। / \(S_6=\frac{4^6-1}{4-1}=1365\). In exams, the form \(\frac{r^n-1}{r-1}\) is useful for (r>1).
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If \(a_n=9\cdot2^{n-1}\), what is the sum of the first (8) terms?
A. (2295)
यहाँ (a=9) और (r=2) है, इसलिए (S_8=9\(2^8-1\)=2295) है। परीक्षा में सामान्य पद से (a) और (r) पहचानें। / Here (a=9) and (r=2), so (S_8=9\(2^8-1\)=2295). In exams, identify (a) and (r) from the general term.
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Which term of the geometric progression \(3,15,75,\ldots\) is (9375)?
B. छठा पद(6)th term
\(3\cdot5^{n-1}=9375\) से \(5^{n-1}=3125=5^5\), इसलिए (n=6) है। परीक्षा में पद संख्या के लिए घात की तुलना करें। / From \(3\cdot5^{n-1}=9375\), \(5^{n-1}=3125=5^5\), so (n=6). In exams, compare powers to find the term number.
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What is the sum of the first (7) terms of the geometric progression \(6,18,54,\ldots\)?
A. (6558)
(S_7=\frac{6\(3^7-1\)}{3-1}=6558) है। परीक्षा में पहले \(3^7\) का मान निकालें। / (S_7=\frac{6\(3^7-1\)}{3-1}=6558). In exams, first calculate the value of \(3^7\).
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If the fifth term of the geometric progression \(x,,5x,,25x,\ldots\) is (3125), what is (x)?
D. (5)
पाँचवाँ पद \(x\cdot5^4=625x=3125\) है, इसलिए (x=5) है। परीक्षा में बीजीय पहले पद को भी \(ar^{n-1}\) में रखें। / The fifth term is \(x\cdot5^4=625x=3125\), so (x=5). In exams, put the algebraic first term in \(ar^{n-1}\) too.
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How many terms are there up to (4096) in the geometric progression \(4,16,64,\ldots\)?
B. (6)
\(4\cdot4^{n-1}=4096\) से \(4^n=4096=4^6\), इसलिए (n=6) है। परीक्षा में अंतिम पद को सामान्य पद के बराबर रखें। / \(4\cdot4^{n-1}=4096\) gives \(4^n=4096=4^6\), so (n=6). In exams, equate the last term to the general term.
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If the first (4) terms of a geometric progression are (7,21,63,189), what is \(S_4\)?
A. (280)
योग (7+21+63+189=280) है। परीक्षा में छोटे (n) के लिए सीधे जोड़ना भी तेज होता है। / The sum is (7+21+63+189=280). In exams, direct addition is also fast for small (n).
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Which term of the geometric progression \(2,8,32,\ldots\) is (8192)?
B. सातवाँ पद(7)th term
\(2\cdot4^{n-1}=8192\) से \(4^{n-1}=4096=4^6\), इसलिए (n=7) है। परीक्षा में घात बराबर करके पद संख्या निकालें। / From \(2\cdot4^{n-1}=8192\), \(4^{n-1}=4096=4^6\), so (n=7). In exams, equate powers to find the term number.
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What is the sum of the first (4) terms of the geometric progression \(5,30,180,1080,\ldots\)?
B. (1295)
पहले चार पदों का योग (5+30+180+1080=1295) है। परीक्षा में पद कम हों तो सीधे जोड़ना सुरक्षित है। / The sum of the first four terms is (5+30+180+1080=1295). In exams, direct addition is safe when the number of terms is small.
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In the geometric progression \(486,162,54,\ldots\), which term is (2)?
B. छठा पद(6)th term
हर बार \(\frac{1}{3}\) से गुणा होता है, इसलिए पद (486,162,54,18,6,2) हैं। परीक्षा में घटती GP को क्रम से भी जाँच सकते हैं। / Each term is multiplied by \(\frac{1}{3}\), so the terms are (486,162,54,18,6,2). In exams, you can also check a decreasing GP in order.
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If \(a_n=405\left(\frac{1}{3}\right)^{n-1}\), which term will be equal to (5)?
C. पाँचवाँ पद(5)th term
\(405\left(\frac{1}{3}\right)^{n-1}=5\) से \(\left(\frac{1}{3}\right)^{n-1}=\frac{1}{81}\), इसलिए (n=5) है। परीक्षा में भिन्नों को पहले सरल करें। / From \(405\left(\frac{1}{3}\right)^{n-1}=5\), \(\left(\frac{1}{3}\right)^{n-1}=\frac{1}{81}\), so (n=5). In exams, simplify fractions first.
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What is the sum of the first (9) terms of the geometric progression \(10,20,40,\ldots\)?
A. (5110)
(S_9=10\(2^9-1\)=5110) है। परीक्षा में \(2^9=512\) याद रखना उपयोगी है। / (S_9=10\(2^9-1\)=5110). In exams, remembering \(2^9=512\) is useful.
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If \(a=96\) and \(r=\frac{1}{2}\), what is the sum of the first \(6\) terms?
B. (189)
पद \(96,48,24,12,6,3\) हैं और योग (189) है। परीक्षा में छोटे घटते पदों को सीधे जोड़ना भी आसान है। / The terms are \(96,48,24,12,6,3\), and their sum is \(189\). In exams, direct addition is also easy for small decreasing terms.
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What is the sum of the first (10) terms of the geometric progression \(12,24,48,\ldots\)?
A. (12276)
(S_{10}=12\(2^{10}-1\)=12276) है। परीक्षा में \(2^{10}=1024\) का उपयोग करें। / (S_{10}=12\(2^{10}-1\)=12276). In exams, use \(2^{10}=1024\).
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A. छठा पद(6)th term
हर बार \(\frac{1}{3}\) से गुणा होता है और छठा पद \(\frac{40}{81}\) आता है। परीक्षा में भिन्न पदों को क्रम से भी जाँच सकते हैं। / Each term is multiplied by \(\frac{1}{3}\), and the sixth term is \(\frac{40}{81}\). In exams, fractional terms can also be checked in order.
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How many terms of the geometric progression \(5,15,45,\ldots\) have sum (1820)?
C. (6)
(S_n=\frac{5\(3^n-1\)}{3-1}) और (\frac{5\(3^n-1\)}{2}=1820) से \(3^n=729\), इसलिए (n=6) है। परीक्षा में योग को सरल करके घात पहचानें। / (S_n=\frac{5\(3^n-1\)}{3-1}), and (\frac{5\(3^n-1\)}{2}=1820) gives \(3^n=729\), so (n=6). In exams, simplify the sum and identify the power.
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