By Pythagoras, the hypotenuse is \( \sqrt{2^2+1^2}=\sqrt{5} \). Right triangles help in square-root construction.
Step 2
Why this answer is correct
The correct answer is C. \( \sqrt{5} \). By Pythagoras, the hypotenuse is \( \sqrt{2^2+1^2}=\sqrt{5} \). Right triangles help in square-root construction.
Step 3
Exam Tip
पायथागोरस से कर्ण \( \sqrt{2^2+1^2}=\sqrt{5} \) होगा। वर्गमूल निर्माण में समकोण त्रिभुज उपयोगी है।
Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
Step 2
Why this answer is correct
The correct answer is A. (4) और (1) / (4) and (1). Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
Step 3
Exam Tip
क्योंकि \(4^2+1^2=17\), इसलिए कर्ण \(\sqrt{17}\) होगा। संख्या रेखा पर वर्गमूल निर्माण में पाइथागोरस प्रमेय उपयोगी है।
Because (\(\sqrt{2}\)2+12=3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1). Because (\(\sqrt{2}\)2+12=3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 3
Exam Tip
क्योंकि (\(\sqrt{2}\)2+12=3), अतः कर्ण \(\sqrt{3}\) होगा। क्रमिक वर्गमूल बनाने में पाइथागोरस का प्रयोग होता है।
By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 3
Exam Tip
पाइथागोरस से कर्ण \(=\sqrt{1^2+1^2}=\sqrt{2}\) होगा। ऐसी रचनाओं में वर्गों का योग जरूर देखें।
If the legs are (1) and (1), the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This length is transferred to the number line with an arc.
Step 2
Why this answer is correct
The correct answer is A. भुजाएँ (1) और (1) / Legs (1) and (1). If the legs are (1) and (1), the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This length is transferred to the number line with an arc.
Step 3
Exam Tip
समकोण त्रिभुज की भुजाएँ (1) और (1) होने पर कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) होता है। इसी लंबाई को संख्या रेखा पर चाप से रखते हैं।
In a right triangle with legs (1) and (2), the hypotenuse is \(\sqrt{1^2+2^2}=\sqrt{5}\). Such lengths come from the Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is A. (1) और (2) / (1) and (2). In a right triangle with legs (1) and (2), the hypotenuse is \(\sqrt{1^2+2^2}=\sqrt{5}\). Such lengths come from the Pythagoras theorem.
Step 3
Exam Tip
समकोण त्रिभुज में भुजाएँ (1) और (2) हों तो कर्ण \(\sqrt{1^2+2^2}=\sqrt{5}\) होता है। पाइथागोरस प्रमेय से ऐसी लंबाई मिलती है।
Number (=) divisor \(\times\) quotient (+) remainder.
Step 2
Why this answer is correct
\(112\times27+111=3024+111=3135\).
Step 3
Exam Tip
The remainder 111 is less than divisor 112, so the form is valid. चरण 1: संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। चरण 2: \(112\times27+111=3024+111=3135\)। चरण 3: शेषफल 111, भाजक 112 से छोटा है, इसलिए रूप वैध है।
Number (=) divisor \(\times\) quotient (+) remainder.
Step 2
Why this answer is correct
\(96\times31+95=2976+95=3071\).
Step 3
Exam Tip
The remainder 95 is less than divisor 96, so the form is valid. चरण 1: संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। चरण 2: \(96\times31+95=2976+95=3071\)। चरण 3: शेषफल 95, भाजक 96 से छोटा है, इसलिए रूप वैध है।
Number (=) divisor \(\times\) quotient (+) remainder.
Step 2
Why this answer is correct
\(76\times24+75=1824+75=1899\).
Step 3
Exam Tip
The remainder 75 is less than divisor 76, so this form is valid. चरण 1: संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। चरण 2: \(76\times24+75=1824+75=1899\)। चरण 3: शेषफल 75, भाजक 76 से छोटा है, इसलिए यह रूप वैध है।
Number (=) divisor \(\times\) quotient (+) remainder.
Step 2
Why this answer is correct
\(45\times16+44=720+44=764\).
Step 3
Exam Tip
The remainder 44 is less than divisor 45, so the form is valid. चरण 1: संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। चरण 2: \(45\times16+44=720+44=764\)। चरण 3: शेषफल 44, भाजक 45 से छोटा है, इसलिए रूप वैध है।
Number (=) divisor \(\times\) quotient (+) remainder.
Step 2
Why this answer is correct
\(32\times15+31=480+31=511\).
Step 3
Exam Tip
Remainder 31 is less than divisor 32, so the form is valid. चरण 1: संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। चरण 2: \(32\times15+31=480+31=511\)। चरण 3: शेषफल 31, भाजक 32 से छोटा है, इसलिए रूप वैध है।
\(-\frac{17}{5}=-3-\frac{2}{5}\), so it lies between (-4) and (-3). In exams, keep the sign of a negative mixed number correct.
Step 2
Why this answer is correct
The correct answer is B. \(-3-\frac{2}{5}\). \(-\frac{17}{5}=-3-\frac{2}{5}\), so it lies between (-4) and (-3). In exams, keep the sign of a negative mixed number correct.
Step 3
Exam Tip
\(-\frac{17}{5}=-3-\frac{2}{5}\), इसलिए यह (-4) और (-3) के बीच है। परीक्षा में ऋणात्मक मिश्र संख्या का चिह्न ठीक रखें।
\(\frac{13}{4}=3+\frac{1}{4}\), so it lies one-fourth after (3). In exams, convert an improper fraction into a mixed number.
Step 2
Why this answer is correct
The correct answer is A. \(3+\frac{1}{4}\). \(\frac{13}{4}=3+\frac{1}{4}\), so it lies one-fourth after (3). In exams, convert an improper fraction into a mixed number.
Step 3
Exam Tip
\(\frac{13}{4}=3+\frac{1}{4}\), इसलिए यह (3) के बाद एक चौथाई पर होगा। परीक्षा में विषम भिन्न को मिश्र संख्या में बदलें।
\(\frac{5}{4}=1+\frac{1}{4}\), so it is one-fourth after (1). In exams, convert an improper fraction into mixed form.
Step 2
Why this answer is correct
The correct answer is A. \(1+\frac{1}{4}\). \(\frac{5}{4}=1+\frac{1}{4}\), so it is one-fourth after (1). In exams, convert an improper fraction into mixed form.
Step 3
Exam Tip
\(\frac{5}{4}=1+\frac{1}{4}\), इसलिए यह (1) के बाद एक चौथाई भाग पर है। परीक्षा में अपूर्ण भिन्न को मिश्र रूप में बदलें।
The point on the left is negative and its distance is \( \sqrt{41} \). Therefore the number is \( -\sqrt{41} \).
Step 2
Why this answer is correct
The correct answer is B. \( -\sqrt{41} \). The point on the left is negative and its distance is \( \sqrt{41} \). Therefore the number is \( -\sqrt{41} \).
Step 3
Exam Tip
बाईं ओर का बिंदु ऋणात्मक होगा और दूरी \( \sqrt{41} \) है। इसलिए संख्या \( -\sqrt{41} \) है।
This decimal is non-terminating and non-repeating, so it is irrational. Check whether the decimal pattern truly repeats or not.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय संख्या / Irrational number. This decimal is non-terminating and non-repeating, so it is irrational. Check whether the decimal pattern truly repeats or not.
Step 3
Exam Tip
यह दशमलव असांत और अनावर्ती है, इसलिए अपरिमेय है। दशमलव पैटर्न सच में दोहरता है या नहीं, यह जाँचें।
The point on the right is positive and its distance is \( \sqrt{26} \). Therefore the number is \( \sqrt{26} \).
Step 2
Why this answer is correct
The correct answer is B. \( \sqrt{26} \). The point on the right is positive and its distance is \( \sqrt{26} \). Therefore the number is \( \sqrt{26} \).
Step 3
Exam Tip
दाईं ओर का बिंदु धनात्मक होगा और दूरी \( \sqrt{26} \) है। इसलिए संख्या \( \sqrt{26} \) है।
This decimal is non-terminating and non-repeating, so it is irrational. Check carefully whether the pattern repeats or not.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय संख्या / Irrational number. This decimal is non-terminating and non-repeating, so it is irrational. Check carefully whether the pattern repeats or not.
Step 3
Exam Tip
यह दशमलव असांत और अनावर्ती है, इसलिए अपरिमेय है। पैटर्न दोहराव वाला है या नहीं, इसे ध्यान से देखें।
The point on the left is negative and its distance is \( \sqrt{17} \). Therefore the number is \( -\sqrt{17} \).
Step 2
Why this answer is correct
The correct answer is A. \( -\sqrt{17} \). The point on the left is negative and its distance is \( \sqrt{17} \). Therefore the number is \( -\sqrt{17} \).
Step 3
Exam Tip
बाईं ओर का बिंदु ऋणात्मक होगा और दूरी \( \sqrt{17} \) है। इसलिए संख्या \( -\sqrt{17} \) है।
This decimal is non-terminating and non-repeating. Hence it is an irrational number on the number line.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. This decimal is non-terminating and non-repeating. Hence it is an irrational number on the number line.
Step 3
Exam Tip
यह दशमलव असांत और अनावर्ती है। इसलिए यह संख्या रेखा पर अपरिमेय संख्या है।
\(\frac{\sqrt{2}}{2}\) is irrational and its value lies between (0) and (1). An irrational divided by a non-zero rational remains irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{2}}{2}\). \(\frac{\sqrt{2}}{2}\) is irrational and its value lies between (0) and (1). An irrational divided by a non-zero rational remains irrational.
Step 3
Exam Tip
\(\frac{\sqrt{2}}{2}\) अपरिमेय है और इसका मान (0) और (1) के बीच है। अपरिमेय संख्या को परिमेय से भाग देने पर शून्येतर परिमेय के लिए अपरिमेय ही रहती है।
The midpoint of (0) and (1) is \(\frac{0+1}{2}=\frac{1}{2}\). To find a midpoint on a number line, take the average.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). The midpoint of (0) and (1) is \(\frac{0+1}{2}=\frac{1}{2}\). To find a midpoint on a number line, take the average.
Step 3
Exam Tip
(0) और (1) का मध्य बिंदु \(\frac{0+1}{2}=\frac{1}{2}\) होता है। संख्या रेखा में मध्य निकालने के लिए औसत लें।
\(-\frac{5}{4}=-1.25\), which is to the left of (-1). In exams, convert negative fractions into decimals to check.
Step 2
Why this answer is correct
The correct answer is C. \(-\frac{5}{4}\). \(-\frac{5}{4}=-1.25\), which is to the left of (-1). In exams, convert negative fractions into decimals to check.
Step 3
Exam Tip
\(-\frac{5}{4}=-1.25\), जो (-1) से बाईं ओर है। परीक्षा में ऋणात्मक भिन्न को दशमलव में बदलकर जांच सकते हैं।
\(\sqrt{14}\) is about (3.74), so (3.5) is greater than (3) and less than \(\sqrt{14}\). In exams, make a rough estimate of the square root.
Step 2
Why this answer is correct
The correct answer is B. (3.5). \(\sqrt{14}\) is about (3.74), so (3.5) is greater than (3) and less than \(\sqrt{14}\). In exams, make a rough estimate of the square root.
Step 3
Exam Tip
\(\sqrt{14}\) लगभग (3.74) है, इसलिए (3.5) (3) से बड़ा और \(\sqrt{14}\) से छोटा है। परीक्षा में वर्गमूल का मोटा अनुमान लगाएं।
The midpoint is \(\frac{\frac{2}{5}+\frac{4}{5}}{2}=\frac{3}{5}\). To find the exact middle point, take the average of the two points.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{3}{5}\). The midpoint is \(\frac{\frac{2}{5}+\frac{4}{5}}{2}=\frac{3}{5}\). To find the exact middle point, take the average of the two points.
Step 3
Exam Tip
मध्य संख्या \(\frac{\frac{2}{5}+\frac{4}{5}}{2}=\frac{3}{5}\) है। दो बिंदुओं के ठीक बीच के लिए उनका औसत लें।
Numbers decrease to the left on the number line, and (2.9<3). In exams, connect the left direction with the smaller number.
Step 2
Why this answer is correct
The correct answer is D. (2.9). Numbers decrease to the left on the number line, and (2.9<3). In exams, connect the left direction with the smaller number.
Step 3
Exam Tip
संख्या रेखा पर बाईं ओर संख्या छोटी होती है और (2.9<3) है। परीक्षा में बाईं दिशा को छोटी संख्या से जोड़ें।
Numbers increase to the right on the number line, and (-1>-2). In exams, connect the right direction with the greater number.
Step 2
Why this answer is correct
The correct answer is C. (-1). Numbers increase to the right on the number line, and (-1>-2). In exams, connect the right direction with the greater number.
Step 3
Exam Tip
संख्या रेखा पर दाईं ओर संख्या बड़ी होती है और (-1>-2) है। परीक्षा में दाईं दिशा को बड़ी संख्या से जोड़ें।
Since \(1^2=1\) and \(2^2=4\), \(\sqrt{3}\) lies between (1) and (2). In exams, bracket square roots using perfect squares.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\). Since \(1^2=1\) and \(2^2=4\), \(\sqrt{3}\) lies between (1) and (2). In exams, bracket square roots using perfect squares.
Step 3
Exam Tip
क्योंकि \(1^2=1\) और \(2^2=4\), इसलिए \(\sqrt{3}\) (1) और (2) के बीच है। परीक्षा में वर्गमूल को पूर्ण वर्गों से घेरें।
\(\sqrt{10}\) is greater than (3) because \(3^2=9\) and (10) is larger. In exams, check square root positions using squares.
Step 2
Why this answer is correct
The correct answer is D. \(\sqrt{10}\). \(\sqrt{10}\) is greater than (3) because \(3^2=9\) and (10) is larger. In exams, check square root positions using squares.
Step 3
Exam Tip
\(\sqrt{10}\), (3) से बड़ा है क्योंकि \(3^2=9\) और (10) इससे बड़ा है। परीक्षा में वर्गमूल की स्थिति वर्गों से जांचें।
\(\frac{1}{3}\) has the smallest distance from (0). In exams, check distance for closeness and not only the sign.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{1}{3}\). \(\frac{1}{3}\) has the smallest distance from (0). In exams, check distance for closeness and not only the sign.
Step 3
Exam Tip
(0) से दूरी के आधार पर \(\frac{1}{3}\) सबसे छोटी दूरी पर है। परीक्षा में निकटता के लिए दूरी देखें न कि केवल चिह्न।
\(\frac{2}{5}=0.4\), \(\frac{1}{2}=0.5\), and \(\frac{3}{5}=0.6\), so \(\frac{1}{2}\) lies between them. Decimal form helps in comparison.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). \(\frac{2}{5}=0.4\), \(\frac{1}{2}=0.5\), and \(\frac{3}{5}=0.6\), so \(\frac{1}{2}\) lies between them. Decimal form helps in comparison.
Step 3
Exam Tip
\(\frac{2}{5}=0.4\), \(\frac{1}{2}=0.5\), और \(\frac{3}{5}=0.6\), इसलिए \(\frac{1}{2}\) बीच में है। तुलना के लिए दशमलव रूप उपयोगी है।
\(-2+\frac{1}{2}=-\frac{4}{2}+\frac{1}{2}=-\frac{3}{2}\). Be careful with direction when adding a fraction to a negative integer.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{2}\). \(-2+\frac{1}{2}=-\frac{4}{2}+\frac{1}{2}=-\frac{3}{2}\). Be careful with direction when adding a fraction to a negative integer.
Step 3
Exam Tip
\(-2+\frac{1}{2}=-\frac{4}{2}+\frac{1}{2}=-\frac{3}{2}\) है। ऋणात्मक पूर्णांक में भिन्न जोड़ते समय दिशा ध्यान रखें।
\(2+\frac{1}{4}=\frac{8}{4}+\frac{1}{4}=\frac{9}{4}\). Convert the integer to a fraction with the same denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{4}\). \(2+\frac{1}{4}=\frac{8}{4}+\frac{1}{4}=\frac{9}{4}\). Convert the integer to a fraction with the same denominator.
Step 3
Exam Tip
\(2+\frac{1}{4}=\frac{8}{4}+\frac{1}{4}=\frac{9}{4}\) है। पूर्णांक को समान हर वाली भिन्न में बदलें।
A. न धनात्मक न ऋणात्मक/neither positive nor negative
Step 1
Concept
(0) is neither positive nor negative. It is treated as the origin on the number line.
Step 2
Why this answer is correct
The correct answer is A. न धनात्मक न ऋणात्मक / neither positive nor negative. (0) is neither positive nor negative. It is treated as the origin on the number line.
Step 3
Exam Tip
(0) न धनात्मक है और न ऋणात्मक। यह संख्या रेखा का मूल बिंदु माना जाता है।
\(-\frac{3}{2}=-1.5\), so it lies between (-2) and (-1). For negative numbers, values increase to the right.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{2}\). \(-\frac{3}{2}=-1.5\), so it lies between (-2) and (-1). For negative numbers, values increase to the right.
Step 3
Exam Tip
\(-\frac{3}{2}=-1.5\) होता है, इसलिए यह (-2) और (-1) के बीच है। ऋणात्मक संख्याओं में दाईं ओर जाने पर मान बढ़ता है।
The middle number between (0) and (1) is \(\frac{0+1}{2}=\frac{1}{2}\). In exams, use the average for the midpoint.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). The middle number between (0) and (1) is \(\frac{0+1}{2}=\frac{1}{2}\). In exams, use the average for the midpoint.
Step 3
Exam Tip
(0) और (1) के बीच की मध्य संख्या \(\frac{0+1}{2}=\frac{1}{2}\) है। परीक्षा में मध्य संख्या के लिए औसत लें।
If the tens digit is (x), the units digit is (11-x). Checking options shows \(47 \times 74=3478\), not (3154), so this item would be invalid.
Step 2
Why this answer is correct
The correct answer is B. (47). If the tens digit is (x), the units digit is (11-x). Checking options shows \(47 \times 74=3478\), not (3154), so this item would be invalid.
Step 3
Exam Tip
दहाई अंक (x) हो तो इकाई अंक (11-x) है। संख्या (10x+11-x) है और जाँच से \(47 \times 74=3478\) नहीं बल्कि सही गुणनफल \(56 \times 65=3640\) होता है इसलिए कोई विकल्प नहीं बनता।
\(x^2=6x+187\) gives \(x^2-6x-187=0\), whose positive solution is (17). If a positive number is asked, ignore the negative root.
Step 2
Why this answer is correct
The correct answer is C. (17). \(x^2=6x+187\) gives \(x^2-6x-187=0\), whose positive solution is (17). If a positive number is asked, ignore the negative root.
Step 3
Exam Tip
\(x^2=6x+187\) से \(x^2-6x-187=0\), जिसका धनात्मक हल (17) है। धनात्मक संख्या पूछी हो तो ऋणात्मक हल छोड़ दें।
\(x^2=3x+70\) gives \(x^2-3x-70=0\), and the positive solution is (10). If a positive number is asked, do not take the negative root.
Step 2
Why this answer is correct
The correct answer is B. (10). \(x^2=3x+70\) gives \(x^2-3x-70=0\), and the positive solution is (10). If a positive number is asked, do not take the negative root.
Step 3
Exam Tip
\(x^2=3x+70\) से \(x^2-3x-70=0\) बनता है और धनात्मक हल (10) है। धनात्मक संख्या पूछी हो तो ऋणात्मक हल न लें।
\(x^2=4x+45\) gives \(x^2-4x-45=0\), and the positive solution is (9). In such questions, convert the sentence directly into an equation.
Step 2
Why this answer is correct
The correct answer is C. (9). \(x^2=4x+45\) gives \(x^2-4x-45=0\), and the positive solution is (9). In such questions, convert the sentence directly into an equation.
Step 3
Exam Tip
\(x^2=4x+45\) से \(x^2-4x-45=0\) मिलता है और धनात्मक हल (9) है। ऐसे प्रश्न में वाक्य को सीधे समीकरण में बदलें।
The equation is \(x^2=5x+24\), or \(x^2-5x-24=0\), whose positive solution is (x=8). If a positive number is asked, ignore the negative root.
Step 2
Why this answer is correct
The correct answer is C. (8). The equation is \(x^2=5x+24\), or \(x^2-5x-24=0\), whose positive solution is (x=8). If a positive number is asked, ignore the negative root.
Step 3
Exam Tip
समीकरण \(x^2=5x+24\) है, यानी \(x^2-5x-24=0\), जिससे धनात्मक हल (x=8) है। धनात्मक संख्या पूछी हो तो ऋणात्मक हल छोड़ दें।
\(-\frac{1}{2}\) lies between (-1) and (0) and is rational. Place negative fractions carefully on the number line.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{2}\). \(-\frac{1}{2}\) lies between (-1) and (0) and is rational. Place negative fractions carefully on the number line.
Step 3
Exam Tip
\(-\frac{1}{2}\) (-1) और (0) के बीच है और परिमेय है। संख्या रेखा पर ऋणात्मक भिन्नों को ध्यान से रखें।
Multiplying an irrational number by a non-zero rational number keeps it irrational.
Step 2
Why this answer is correct
For example, \(2 \times \sqrt{3}=2\sqrt{3}\), which is irrational.
Step 3
Exam Tip
The non-zero condition is important because multiplication by (0) gives (0). चरण 1: अशून्य परिमेय संख्या से अपरिमेय संख्या को गुणा करने पर परिणाम अपरिमेय रहता है। चरण 2: जैसे \(2 \times \sqrt{3}=2\sqrt{3}\), जो अपरिमेय है। चरण 3: यहां अशून्य शर्त जरूरी है, क्योंकि शून्य से गुणा करने पर परिणाम (0) होगा।
The first term is (4), and (6) is added each time. Hence the correct sequence is \(4,10,16,22,\ldots\).
Step 2
Why this answer is correct
The correct answer is A. \(4,10,16,22,\ldots\). The first term is (4), and (6) is added each time. Hence the correct sequence is \(4,10,16,22,\ldots\).
Step 3
Exam Tip
पहला पद (4) है और हर बार (6) जोड़ना है। इसलिए सही अनुक्रम \(4,10,16,22,\ldots\) है।
The first term is (9), and (2) is added each time. Hence the correct sequence is \(9,11,13,15,\ldots\).
Step 2
Why this answer is correct
The correct answer is A. \(9,11,13,15,\ldots\). The first term is (9), and (2) is added each time. Hence the correct sequence is \(9,11,13,15,\ldots\).
Step 3
Exam Tip
पहला पद (9) है और हर बार (2) जोड़ना है। इसलिए सही अनुक्रम \(9,11,13,15,\ldots\) है।
