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In an AP, if d = –4, n = 7, an=4, then a is

(A) 6

(B) 7

(C) 20

(D) 28

Given

\\a\textsubscript{n} = 4, d = - 4, n = 7 \\Use \ a\textsubscript{n} = a + (n - 1).d\\ a + (7 - 1). (- 4) = 4\\ a +(6).(-4) = 4\\ a - 24 = 4\\ a = 4 + 24\\ a = 28\\

 Hence option D is correct

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The sum of first five multiples of 3 is

(A) 45

(B) 55

(C) 65

(D) 75

\\$Given:- Multiple of 3 is : 3, 6, 9, 12, $ \ldots $ $ \ldots $ .\\$ a\textsubscript{1 }= 3, d = a\textsubscript{2} - a\textsubscript{1}\\ d = 6 - 3\\ d = 3\\ {{S}_{5}}=\frac{5}{2}[2{{a}_{1}}+(5-1)d] \\ =\frac{5}{2}(2\times 3+4\times 3) \\ =\frac{5}{2}(6+12) \\ =\frac{5}{2}\times (18) \\

Hence option A is correct.

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\text{In an AP if a = 1,} a\textsubscript{n} = 20 and S\textsubscript{n} = 399\text{, then n is }

(A) 19

(B) 21

(C) 38

(D) 42

\\$Given : a = 1, a\textsubscript{n} = 20, S\textsubscript{n} = 399\\$ {{S}_{n}}=\frac{n}{2}(a+{{a}_{n}}) \\ 38 = n\\

Hence option C is correct.

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The sum of first 16 terms of the AP: 10, 6, 2... is

(A) –320

(B) 320

(C) –352

(D) –400

\\$Given:AP=10,6,2$ \ldots $ here a\textsubscript{1}=firsttermandd=commondifference \\$ a\textsubscript{1} = 10, a\textsubscript{2}=6\\ d = a\textsubscript{2} - a\textsubscript{1 }\\ d =6 - 10\\ d = - 4\\ {{S}_{16}}=\frac{16}{2}(2a+(16-1)d) \\ = 8(2 \times 10 + 15 \times (-4))\\ = 8(20 - 60)\\ = 8 \times (-40) = - 320\\

Hence option A is the correct answer.

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If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is

(A) 0

(B) 5

(C) 6

(D) 15

\\$Given a=-5,d=2 here a= first term and d= common difference\\$ S\textsubscript{6} = \frac{6}{2}(2a+(6-1)d) \\ = 3(2 \times (-5) + 5 \times 2)\\ = 3(-10 + 10)\\ = 3(0) = 0\\

Hence option A is correct

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The famous mathematician associated with finding the sum of the first 100 natural numbers is

(A) Pythagoras

(B) Newton

(C) Gauss

(D) Euclid

[C]

Famous mathematician gauss find the sum of the first 100 natural numbers.

According to which when he added the first and the last number in the series, the second and the second last number in the series, and so on. He get a sum of 101 and there are 50 pairs exist so the sum of 100 natural numbers is 50×101=5050.

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The 4thterm from the end of the AP: –11, –8, –5, ..., 49 is

(A) 37

(B) 40

(C) 43

(D) 58

\text{Given AP is } -11, -8, -5, \ldots, 49 \\ \text{Here, } a_{1} = -11, \quad a_{2} = -8, \quad a_{n} = 49 \\ d = a_{2} - a_{1} \\ = -8 - (-11) \\ = -8 + 11 \\ = 3 \\ \text{We know that } a_{n} = a_{1} + (n - 1) \cdot d \\ \text{Put } 49 = -11 + (n - 1)(3) \\ 49 = -11 + 3n - 3 \\ 49 = -14 + 3n \\ 49 + 14 = 3n \\ 63 = 3n \\ n = \frac{63}{3} \\ n = 21 \\ a_{n} = a_{1} + (n - 1) \cdot d \\ \text{Put } n = 21 - 3 = 18 \\ a_{18} = -11 + (18 - 1) \times 3 \\ a_{18} = -11 + 17 \times 3 \\ a_{18} = -11 + 51 \\ a_{18} = 40 \\ \text{40 is the 4th last term from end.}

Hence option B is correct

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If 7 times the 7thterm of an AP is equal to 11 times its 11th term, then its 18thterm will be

(A) 7

(B) 11

(C) 18

(D) 0

\text{Given } 7 \times a_{7} = 11 \times a_{11} \\ 7 \times (a + 6d) = 11 \times (a + 10d) \quad (\text{Using } a_{n} = a + (n - 1) \times d) \\ 7a + 42d = 11a + 110d \\ 11a - 7a + 110d - 42d = 0 \\ 4a + 68d = 0 \\ \text{Dividing by } 4 \text{, we get} \\ a + 17d = 0 \quad (1) \\ a_{18} = a + (18 - 1)d \\ = a + 17d \\ = 0 \\

Hence option D is correct.

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Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8. Then the difference between their 4th terms is

(A) –1

(B) – 8

(C) 7

(D) –9

\text{Let the first AP be } a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots \ldots \text{ (1)} \\ \text{Let the second AP be } b_{1}, b_{2}, b_{3}, \ldots \ldots \ldots \ldots \text{ (2)} \\ \text{Given that } a_{1} = -1, \quad b_{1} = -8 \\ \text{Difference between their 4th terms is} \\ a_{4} - b_{4} = (a_{1} + 3d) - (b_{1} + 3d) \\ \text{Given that the common difference of both APs is equal} \\ a_{4} - b_{4} = -1 - (-8) \\ = -1 + 8 \\ = 7 \\

 Hence option (C) is correct

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What is the common difference of an AP in which a18 - a14 = 32?

(A)  8

(B) – 8

(C) – 4

(D)   4

Given,

\\a\textsubscript{18 }- a\textsubscript{14} = 32 \\ $We know that a\textsubscript{n} $= a + (n - 1).d\\ $Put n $= 18\\ a\textsubscript{18} = a + (18 - 1).d\\ a\textsubscript{18} = 9 + 17d$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ ..(1)\\ n = 14\\ a\textsubscript{14} = a + (14 - 1).d\\ a\textsubscript{14} = 9 + 13d$ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ $ \ldots $ .(2)\\ $subtract eq (2) from eq(1)\\$ a\textsubscript{18}- a\textsubscript{14} = 32\\ (a + 17d) - (a + 13d) = 32\\ a + 17d - a - 13d = 32\\ 4d = 32\\ d = \frac{32}{4} \\ d=8 \\

Hence common difference is 8.

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