How many terms of ap 27 24 21
WebAnswer: Solution: The sum of an AP's n terms is given by S n = n/2 {2a + (n-1)d} where, the AP's first term is 'a 'and the common difference is 'd'. Given A.P. is 27, 24, 21. . . We … Web11 apr. 2024 · $ \Rightarrow d = \left( {21 - 24} \right) = \left( {18 - 21} \right) = - 3$ Now we have to find out the number of terms in the A.P if the sum of an A.P is 78. As we know that the sum (Sn) of an A.P is given as
How many terms of ap 27 24 21
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Web7 sep. 2024 · The term of the A.P. which is 0 is 10th. To answer this question, we need to follow the following steps: As we know the nth term of an AP that has a = first term, d = common difference and n = number of terms till the nth term is given by using the formula: Now, according to the question, We have an A.P. 27, 24, 21. where the first term or a = 27. Web13 okt. 2024 · Step-by-step explanation: The given AP is,27,24,21,.... Here first term= a=27 common difference=d=24-27=-3 Let the nth term of the AP=0 => a+ (n-1)d=0 =>27+ (n-1) (-3)=0 => (n-1)= (-27)/ (-3) => n-1=9 => n=9+1 => n=10 So, the 10th term of the AP is 0. hope it will help you. Advertisement dawin32arpit Answer: 10 please mark as brainlist answer
Web14 jan. 2024 · Best answer Solution: Given AP is -6, -11/2, -5… Here a = –6, d = (–11/2) + 6 = 1/2 Sum of n terms of AP, Sn = – 25 Using sum of n term of AP formula ⇒ n2 – 25n + 100 = 0 ⇒ n2 – 20n – 5n + 100 = 0 ⇒ n (n – 20) – 5 (n – 20) = 0 ⇒ (n – 20) (n – 5) = 0 ⇒ (n – 20) = 0 or (n – 5) = 0 ∴ n = 20 or n = 5 ← Prev Question Next Question → Web28 mrt. 2024 · We know that the sum of all terms of an A.P. is given by, S n = n 2 [ 2 a + ( n − 1) d], ……… (i) this can also be written as: S n = n 2 [ a + l], ………. (ii) here l = last term = a + (n-1)d. Here given that, a = 27 and S n = 0 We know that d = common difference = a 2 − a 1 = 24 − 27 = − 3. Putting these values in equation (i), we get
Web12 aug. 2024 · Find an answer to your question how many terms of the ap 27 24 21 .....should be taken so that their sum is zero. hahshha hahshha 12.08.2024 Math Primary … Web10 okt. 2024 · We have to find the number of terms that must be taken so that their sum is 180. Solution: Let the number of terms be n. First term ( a) = 45 Common difference ( d) = 39 − 45 = − 6 We know that, S n = n 2 [ 2 a + ( n − 1) d] ⇒ 180 = n 2 [ 2 × 45 + ( n − 1) × ( − 6)] ⇒ 180 = n 2 [ 90 − 6 n + 6] ⇒ 360 = n ( − 6 n + 96) ⇒ 6 × 60 = 6 ( − n 2 + 16 n)
Web29 mrt. 2024 · Given AP 24, 21, 18,………. Here, a = 24 d = 21 – 24 = –3 Also, given Sum = 78 Sn = 78 We have to find value of n Putting these values in equation Sum = 𝒏/𝟐 [𝟐𝒂+ (𝒏−𝟏)𝒅] …
Web4 mrt. 2024 · How many Terms of the AP 24 21 18 must be taken so that their Sum is 78 Class 10 Maths Chapter 5 Example 13Example 13 : How many terms of the AP : 24, 21, ... fixed navbar covers anchor contentWeb5 apr. 2024 · number of terms in an. A P = n 2 [ 2 a + ( n − 1) d] . Complete step-by-step answer: Series of numbers given is 21, 18, 15 …. Here in this series of AP first term is 21. We can calculate the common difference by subtracting either 21 from 18 or 18 from 15. d = 18 − 21 = 15 − 18 = − 3. So, the common difference is. fixed costs are defined as:Web12 dec. 2016 · How many terms of the AP 27,24,21, should be taken so that their sum is zero - Maths - Arithmetic Progressions fixed asset trade in accountingWeb28 mrt. 2024 · We know that the sum of all terms of an A.P. is given by, S n = n 2 [ 2 a + ( n − 1) d], ……… (i) this can also be written as: S n = n 2 [ a + l], ………. (ii) here l = last … fixed width integerWebHow many terms of the A.P; 24,21,18,... must be taken such that their sum is 78 Easy Solution Verified by Toppr Formula, S n= 2n[2a+(n−1)d] Given, a=24,d=21−24=−3,S n=78 78= 2n[2(24)+(n−1)(−3)] 156=n[48−3n+3] 3n 2−51n+156=0 n 2−17n+52=0 (n−13)(n−4)=0 ∴n=4,13 Was this answer helpful? 0 0 Similar questions fixed a bugWebHow many terms of the AP:24,21,18.... must be taken so that their sum is 78 ? Hard Solution Verified by Toppr Given: 24,21,18,... are in A.P a=24,d=21−24=−3 Sum = 2n[2a+(n−1)d] ⇒78= 2n[2×24+(n−1)(−3)] ⇒156=n[48−3n+3] ⇒156=n[51−3n] ⇒3n 2−51n+156=0 ⇒3n 2−12n−39n+156=0 ⇒3n(n−4)−39(n−4)=0 ⇒(n−4)(3n−39)=0 ∴n=4,n= … fixed assets setup in d365Web25 jul. 2024 · let first term of the AP be a1= -10 second term of the AP be a2=-7 and common difference be d. d=a2-a1 d=-7- (-10) d=-7+10 d=3 so common difference is d=3 we know that, Sum of n terms in the AP is, Sn=n/2 [2a1+ (n-1)d] 104=n/2 [2×-10 + (n-1)3] 104×2=n [-20+3n-3] 208=n [3n-23] 208=3n^2-23n 3n^3-23n-208=0 3n^2-39n+16n-208=0 … fixed ignition timing