If a1 a2 a3 are in ap such that a1+a7+a16 40
WebClick here👆to get an answer to your question ️ If a1, a2, a3, .... are in A.P. such that a4a7 = 32 , then the 13^th of the A.P. is Web14 jul. 2024 · Best answer Given : a4/a7 = 2/3 To find : a6 a8 a 6 a 8 We know, an = a + (n – 1)d Where a is a first term or a1 and d is the common difference and n is any natural number When n = 4 : ∴ a4 = a + (4 – 1)d ⇒ a4 = a + 3d When n = 6 : ∴ a6 = a + (6 – 1)d ⇒ a6 = a + 5d When n = 7 : ∴ a7 = a + (7 – 1)d ⇒ a7 = a + 6d When n = 8 : ∴ a8 = a + (8 – 1)d
If a1 a2 a3 are in ap such that a1+a7+a16 40
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Web12 apr. 2024 · The nth term in AP series is given by the formula below: T n = a + (n – 1)d Where, n = Number of terms a = First term d = Common difference Now, the condition … WebCorrect option is B) Given equation can be written as a 1+(a 1+3d)+(a 1+6d)+....+a 1+15d=147 ⇒6a 1+3d(1+2+..+5)=147 ⇒2a 1+15d=49 Now, a 1+a 6+a 11+a 16=4a 1+5d+10d+15d =4a 1+30d=2(2a 1+15d)=2×49 =98 ∴ Option B is correct. Was this answer helpful? 0 0 Similar questions
Web20 mei 2024 · If a1, a2, a3 are positive numbers forming G.P. such that a5 + a7 = 12 and a4.a6 = 9, then a1a9 + a2a8 + a5 + a7 is equal to asked Feb 1 in Mathematics by … WebA1,A2,A3,....is in AP the write in common terms of a and d. .i.e. A1 = a A2 =a+d A3 =a+2d . . A24 = a+23d summing all we get, 4 (6a+69d) = > 4*225 = 900 perhaps. krunal 52 Points 6 years ago u can also solve in such way, (A1+A24) + (A5+A20) + (A10+A15)=225. THEREFORE 3 (A1+A24)=225 (A1+A24)=75. NOW SUM=24/12 (A1+A24) =900
WebIf 1 a1a2+ 1 a2a3 +⋯+ 1 a4000a4001 =10 and a2+a4000= 50, then A a1a4001 =400 B a1a4001 =401 C a1−a4001 =40 D a1−a4001 =30 Solution The correct options are A … WebCorrect option is C) It is given that a 1,a 2,a 3,.......,a n are in A.P and we know that: T he general form of an Arithmetic Progression is a,a+d,a+2d,a+3d and so on. Thus n th term …
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WebClick here👆to get an answer to your question ️ Let a1, a2, a3, .... be terms of an A.P.If a1 + a2 + .... + apa1 + a2 + ..... + aq = p^2q^2, ... are in A.P. such that a 1 ... Find a Value of … hawkes property groupWeb24 feb. 2024 · The nth term in AP series is given by the formula below: T n = a + (n – 1)d Where, n = Number of terms a = First term d = Common difference Now, the condition given becomes, ⇒ a 1 + (a 1 + 6d) + (a 1 + 15d) = 40 ⇒ 3a 1 + 21d = 40 ---- (1) Now, the sum of the first 15 terms is given as: The sum of the n terms is given by the formula: hawkes property group londonWeb18 nov. 2014 · if a1,a2,a3, a4001 are terms of an AP such that 1/a1a2 +1/a2a3+ +1/a4000a4001 = 10 and a2+a4000=50, then a1-a4001 is equal to (A)20 (B)30 (C)40 (D) none of ... +1/a4000a4001 = 10 and a2+a4000=50, then a1-a4001 is equal to (A)20 (B)30 (C)40 (D) none of these. Share with your friends. Share 14. We know that, ... hawkes racing stableWebSolution The correct option is C 11 41 Step 1: Determine the common difference d An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant. Sum of n terms of AP, is given by S n = n 2 2 a 1 + n - 1 d where a 1 = first term, d = common difference. hawkes publishingWebIf A1, A2, A3, .... an Are in A.P. with Common Difference D, Then the Sum of the Series Sin D [Cosec A1 Cosec A2 + Cosec A1 Cosec A3 + .... + Cosec an − 1 Cosec An] is - Mathematics Shaalaa.com Department of Pre-University Education, Karnataka PUC Karnataka Science Class 11 Textbook Solutions 11069 Important Solutions 5 hawkes racing syndicationsWeb13 okt. 2024 · 0:00 / 7:21 a1,a2,a3...,an are in A.P. such that a1 + a3 + a5 = -12 and a1a2a3=8 then: 2,308 views Oct 13, 2024 10 Dislike Share Save Doubtnut 2.1M subscribers Subscribe To ask... hawkes racing stablesWebIf a1, a2, a3,........ are terms of AP such that a1 + a5 + a10 + a15 + a20 + a24 = 225 , then the sum of first 24 terms is Class 11 >> Applied Mathematics >> Sequences and series >> Arithmetic progression >> If a1, a2, a3,........ are terms of AP s Question hawkes racing melbourne