WebGiven a, b, c are in GP . So b 2 = a c.. (i) a (b 2 + c 2) = a (a c + c 2) = a (a c) + a c 2 = a 2 c + a c = c (a 2 + b 2) a (b 2 + c 2) = c (a 2 + b 2) Hence, the value of a (b 2 + c 2) i s c … WebClick here👆to get an answer to your question ️ If a,b,c are in AP and a,b,d are in GP then a,(a - b),(d - c) are in Solve Study Textbooks Guides Join / Login
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WebIf a, b, c, d are in GP, prove that (b + c) (b + d) = (c + a) (c + d) . Question If a, b, c, d are in GP, prove that (b+c)(b+d)=(c+a)(c+d). Medium Solution Verified by Toppr Was this … WebIf a, b, c are in G.P. and a x = b y = c z, then prove that 1 x + 1 z = 2 y. View More. Related Videos. MATHEMATICS. Watch in App. Explore more. Geometric Progression. Standard XI Mathematics. Solve. Textbooks. Question Papers.
WebSolution: Since a, b, c, are in A.P, ∴ b = 2a+c …(1) Since b, c, d are in G.P, ∴ c2 = bd ... (2) Since c, d, e, are in H.P, ∴ d = c+e2ce ... (3) Putting the values of b, d in (2) from (1), (3), we get c2 = 2a+c ⋅ c+e2ce = c+e(a+c)ce ⇒ (c +e)c2 = (a+ c)ce ⇒ (c +e)c = (a+ c)e ⇒ c2 + ce = ae +ce ⇒ c2 = ae ⇒ a,b,e are in G.P. WebCorrect option is C) Given : a, c, b are in GP. Let the common ratio be r. Then c=ar and b=ar 2. Given line is ax+by+c=0 ∴ax+by=−c Divide by -c throughout −cax+ −cby=1 ( …
WebIf a, b, c are in GP then a+b, 2b, b+c are in A AP B GP C HP D none of these Medium Solution Verified by Toppr Correct option is C) Given b 2=ac Let us consider, a+b1 + … Web18 jan. 2024 · If a, b, c are in GP and loga − log2b, log2b − log3c and log3c − loga are AP, then a, b, c are the lengths of the sides of a triangle which is (A) acute angled (B) obtuse angled (C) right angled (D) no triangle will be formed binomial theorem jee jee mains 1 Answer +1 vote answered Jan 18, 2024 by KumariJuly (53.9k points)
Web8 aug. 2024 · If a, b, c, d, e are in G.P., then e/c equals : (a) d/b (b) c/b (c) b/a (d) d/c bitsat Share It On 1 Answer +1 vote answered Aug 8, 2024 by Aarju (69.8k points) selected Aug 8, 2024 by faiz Best answer Correct option (a) d/b Explanation: Let a, b, c, d, e be in G.P., with common ratio r. ← Prev Question Next Question → Find MCQs & Mock Test
Web8 feb. 2024 · If a b and c are in gp then a-b/b-c is equal to See answer Advertisement sk940178 where, r is the common ratio. Step-by-step explanation: Given that a, b, c are in G.P. Therefore, if the common ratio of the G.P. is r, then b = ar and c = ar². Now, we are asked to determine the value of . food delivery near me 01060WebCorrect option is A) Given, a,b,c are in G.P. Thus, their common ratio must be equal. Let the common ratio be r. Thus, r= ab= bc .... (1) ⇒b 2=ac⇒a= cb 2. Then value of b−ca−b= … elasticsearch with dockerWebThis is the Solution of Question From RD SHARMA book of CLASS 11 CHAPTER SEQUENCES AND SERIES This Question is also available in R S AGGARWAL book of CLASS 1... elasticsearch with kibana docker composeWeb23 okt. 2024 · Correct Answer - C a, b, c, d a, b, c, d are in GP a = a, b = ar, c = ar2, d = ar3 a = a, b = a r, c = a r 2, d = a r 3 b + c = ar + ar2 = ar(1 + r) b + c = a r + a r 2 = a r ( 1 + r) (b + c)2 + a2r2(1 + r2 + 2r) ( b + c) 2 + a 2 r 2 ( 1 + r 2 + 2 r) = a2r2 + a2r4 + 2a2r3 = a 2 r 2 + a 2 r 4 + 2 a 2 r 3 = a ⋅ ar2 + ar ⋅ ar3 + a2r3 + a2r3 = a ⋅ food delivery near farr west utahWebIf a,b,c are in GP, then the equation ax^2 + 2bx + c = 0 and dx^2 + 2ex + f = 0 have a common root, if da , eb , fc are in Question If a,b,c are in GP, then the equation ax … elasticsearch with javaWebIf a, b, c, d are in G.P., then ( (a2 +b2 + c2) (b2+c2+d2)/ (ab+bc +cd)2)= Q. If a,b,c,d are in G.P ., then (ab+bc+cd)2(a2+b2+c2)(b2+c2+d2) = 1736 31 Sequences and Series Report Error A 1 B 2 C 3 D 5 Solution: b = ar,c = ar2,d = ar3 ∴ a2 +b2 +c2 = a2(1+r2 +r4) b2 +c2 +d2 = a2r2 (1+r2 + r4) ab+bc+ cd = a2r(1+ r2 + r4) food delivery near addison ilWebSolution: Since, a,b,c and d are in GP. ∴ b2 = ac,c2 = bd and ab = cd ⇒ bc = ad...(i) Now, (a +b +c +d)2 = (a +b)2 + (c +d)2 +2(a +b)(c+ d) = (a +b)2 +(c +d)2 +2(ac+ ad+ bc +bd) … elasticsearch withaggregations