Here \(r=\frac{1}{5}\), so the sixth term is (625\cdot\left\(\frac{1}{5}\right\)5=\frac{1}{5}). In exams, apply fractional ratios carefully in decreasing GPs.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{1}{5}\). Here \(r=\frac{1}{5}\), so the sixth term is (625\cdot\left\(\frac{1}{5}\right\)5=\frac{1}{5}). In exams, apply fractional ratios carefully in decreasing GPs.
Step 3
Exam Tip
यहाँ \(r=\frac{1}{5}\) है, इसलिए छठा पद (625\cdot\left\(\frac{1}{5}\right\)5=\frac{1}{5}) है। परीक्षा में घटती GP में भिन्न अनुपात सावधानी से लगाएँ।
From \(6\cdot2^{n-1}=1536\), \(2^{n-1}=256=2^8\), so (n=9). In exams, equate powers to find the term number.
Step 2
Why this answer is correct
The correct answer is C. नौवाँ पद / (9)th term. From \(6\cdot2^{n-1}=1536\), \(2^{n-1}=256=2^8\), so (n=9). In exams, equate powers to find the term number.
Step 3
Exam Tip
\(6\cdot2^{n-1}=1536\) से \(2^{n-1}=256=2^8\), इसलिए (n=9) है। परीक्षा में घातों को बराबर करके पद संख्या निकालें।
The first term is (12) and the ratio is (4), so \(a_n=12\cdot4^{n-1}\). In exams, keep (a) and (r) correct in \(ar^{n-1}\).
Step 2
Why this answer is correct
The correct answer is A. \(a_n=12\cdot4^{n-1}\). The first term is (12) and the ratio is (4), so \(a_n=12\cdot4^{n-1}\). In exams, keep (a) and (r) correct in \(ar^{n-1}\).
Step 3
Exam Tip
पहला पद (12) और अनुपात (4) है, इसलिए \(a_n=12\cdot4^{n-1}\) है। परीक्षा में \(ar^{n-1}\) में (a) और (r) सही रखें।
The first five terms are (4,20,100,500,2500), and the sum is (3124). In exams, add terms carefully when the ratio is large.
Step 2
Why this answer is correct
The correct answer is B. (3124). The first five terms are (4,20,100,500,2500), and the sum is (3124). In exams, add terms carefully when the ratio is large.
Step 3
Exam Tip
पहले पाँच पद (4,20,100,500,2500) हैं और योग (3124) है। परीक्षा में बड़े अनुपात में पदों को सावधानी से जोड़ें।
The first term is (100) and the ratio is \(\frac{1}{2}\), so the correct rule is \(100\cdot\left(\frac{1}{2}\right)^{n-1}\). In exams, write the fractional ratio in a decreasing GP.
Step 2
Why this answer is correct
The correct answer is B. \(a_n=100\cdot\left(\frac{1}{2}\right)^{n-1}\). The first term is (100) and the ratio is \(\frac{1}{2}\), so the correct rule is \(100\cdot\left(\frac{1}{2}\right)^{n-1}\). In exams, write the fractional ratio in a decreasing GP.
Step 3
Exam Tip
पहला पद (100) और अनुपात \(\frac{1}{2}\) है, इसलिए सही नियम \(100\cdot\left(\frac{1}{2}\right)^{n-1}\) है। परीक्षा में घटती GP में भिन्न अनुपात लिखें।
A. सही क्योंकि \(S_4=340\)/True because \(S_4=340\)
Step 1
Concept
(S_4=4\(4^4-1\)/(4-1)=340), so the statement is true. In exams, verify the statement using the formula.
Step 2
Why this answer is correct
The correct answer is A. सही क्योंकि \(S_4=340\) / True because \(S_4=340\). (S_4=4\(4^4-1\)/(4-1)=340), so the statement is true. In exams, verify the statement using the formula.
Step 3
Exam Tip
(S_4=4\(4^4-1\)/(4-1)=340) है, इसलिए कथन सही है। परीक्षा में कथन की जाँच सूत्र से करें।
From \(4\cdot3^{n-1}=972\), \(3^{n-1}=243=3^5\), so (n=6). In exams, compare powers to find the term number.
Step 2
Why this answer is correct
The correct answer is C. छठा पद / (6)th term. From \(4\cdot3^{n-1}=972\), \(3^{n-1}=243=3^5\), so (n=6). In exams, compare powers to find the term number.
Step 3
Exam Tip
\(4\cdot3^{n-1}=972\) से \(3^{n-1}=243=3^5\), इसलिए (n=6) है। परीक्षा में पद संख्या के लिए घात की तुलना करें।
From (1024\left\(\frac{1}{2}\right\)^{n-1}=2), \(2^{10-(n-1)}=2^1\), so (n=10). In exams, count decreasing powers carefully.
Step 2
Why this answer is correct
The correct answer is C. दसवाँ पद / (10)th term. From (1024\left\(\frac{1}{2}\right\)^{n-1}=2), \(2^{10-(n-1)}=2^1\), so (n=10). In exams, count decreasing powers carefully.
Step 3
Exam Tip
(1024\left\(\frac{1}{2}\right\)^{n-1}=2) से \(2^{10-(n-1)}=2^1\), इसलिए (n=10) है। परीक्षा में घटती घातों को सावधानी से गिनें।
From \(10\cdot4^{n-1}=10240\), \(4^{n-1}=1024=4^5\), so (n=6). In exams, equate powers to find the term number.
Step 2
Why this answer is correct
The correct answer is A. छठा पद / (6)th term. From \(10\cdot4^{n-1}=10240\), \(4^{n-1}=1024=4^5\), so (n=6). In exams, equate powers to find the term number.
Step 3
Exam Tip
\(10\cdot4^{n-1}=10240\) से \(4^{n-1}=1024=4^5\), इसलिए (n=6) है। परीक्षा में घात बराबर करके पद संख्या निकालें।
The sum of the first four terms is (2+14+98+686=800). In exams, direct addition is safe when the number of terms is small.
Step 2
Why this answer is correct
The correct answer is B. (800). The sum of the first four terms is (2+14+98+686=800). In exams, direct addition is safe when the number of terms is small.
Step 3
Exam Tip
पहले चार पदों का योग (2+14+98+686=800) है। परीक्षा में पद कम हों तो सीधे जोड़ना सुरक्षित है।
Each term is multiplied by \(\frac{1}{3}\), so the terms are (405,135,45,15,5). In exams, you can also check a decreasing GP in order.
Step 2
Why this answer is correct
The correct answer is B. पाँचवाँ पद / (5)th term. Each term is multiplied by \(\frac{1}{3}\), so the terms are (405,135,45,15,5). In exams, you can also check a decreasing GP in order.
Step 3
Exam Tip
हर बार \(\frac{1}{3}\) से गुणा होता है, इसलिए पद (405,135,45,15,5) हैं। परीक्षा में घटती GP को क्रम से भी जाँच सकते हैं।
From \(128\left(\frac{1}{2}\right)^{n-1}=8\), \(\left(\frac{1}{2}\right)^{n-1}=\frac{1}{16}\), so (n=5). In exams, simplify fractions first.
Step 2
Why this answer is correct
The correct answer is B. पाँचवाँ पद / (5)th term. From \(128\left(\frac{1}{2}\right)^{n-1}=8\), \(\left(\frac{1}{2}\right)^{n-1}=\frac{1}{16}\), so (n=5). In exams, simplify fractions first.
Step 3
Exam Tip
\(128\left(\frac{1}{2}\right)^{n-1}=8\) से \(\left(\frac{1}{2}\right)^{n-1}=\frac{1}{16}\), इसलिए (n=5) है। परीक्षा में भिन्नों को पहले सरल करें।
The terms are (162,54,18,6,2), and the sum is (242). In exams, direct addition is also easy for small decreasing terms.
Step 2
Why this answer is correct
The correct answer is B. (242). The terms are (162,54,18,6,2), and the sum is (242). In exams, direct addition is also easy for small decreasing terms.
Step 3
Exam Tip
पद (162,54,18,6,2) हैं और योग (242) है। परीक्षा में छोटे घटते पदों को सीधे जोड़ना भी आसान है।
Each term is multiplied by \(\frac{1}{2}\), and the seventh term is \(\frac{3}{2}\). In exams, fractional terms can also be checked in order.
Step 2
Why this answer is correct
The correct answer is B. सातवाँ पद / (7)th term. Each term is multiplied by \(\frac{1}{2}\), and the seventh term is \(\frac{3}{2}\). In exams, fractional terms can also be checked in order.
Step 3
Exam Tip
हर बार \(\frac{1}{2}\) से गुणा होता है और सातवाँ पद \(\frac{3}{2}\) आता है। परीक्षा में भिन्न पदों को क्रम से भी जाँच सकते हैं।
(S_n=\frac{4\(3^n-1\)}{3-1}=2\(3^n-1\)), and (2\(3^n-1\)=484) gives (n=5). In exams, simplify the sum and identify the power.
Step 2
Why this answer is correct
The correct answer is B. (5). (S_n=\frac{4\(3^n-1\)}{3-1}=2\(3^n-1\)), and (2\(3^n-1\)=484) gives (n=5). In exams, simplify the sum and identify the power.
Step 3
Exam Tip
(S_n=\frac{4\(3^n-1\)}{3-1}=2\(3^n-1\)) और (2\(3^n-1\)=484) से (n=5) है। परीक्षा में योग को सरल करके घात पहचानें।
