If a b c are in ap and a2 b2 c2 are in gp
Web7 feb. 2024 · Suppose a, b, c are in AP and a 2, b 2, c 2 are in GP. If a < b < c and a + b + c = 3/2 then the value of a is sequences and series jee jee mains 1 Answer +1 vote answered Feb 7, 2024 by Aksat (69.7k points) selected Feb 7, 2024 by Vikash Kumar Best answer The correct option is (d) 1/2 - 1/√2 Explanation: Since. a. h. r are in AP. WebSolution The correct option is A a = b = c Explanation for correct option: Progression: Given, a, b, c are in Arithmetic Progression 2 b = a + c ⇒ b - a = c - b ⇒ a - b = b - c Also, a 2, b 2 & c 2 are in ‘Harmonic progression’, then
If a b c are in ap and a2 b2 c2 are in gp
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WebA Série A1 do Campeonato Paulista de Futebol de 2024, ou Paulistão Sicredi 2024 por motivos de patrocínio, foi a 122ª edição da principal divisão do futebol paulista. [ 2] Foi realizada e organizada pela Federação Paulista de Futebol e disputada por 16 clubes. O campeão da competição foi o Palmeiras, que chegou ao segundo título ... Web4^yÔ^äk:q LDçö P‹ x´Â‰ö”çÇ8$§=ÆÅ Ä F+FC«¶«" W W몬P>#¾¢ ß/ !EB] ¶äÆRÏ }O}u¸Í uÛÓÜœo5&>•v~ó ãÆ3윆/Jº m( Õá! !2h¶t>Xp¯Û³r¬qp[ Ú9+òîéŒóð¶6,þøÕ–}Òü¢§2üŒ¨šæ ®–ºÅmúöpÒ— á [Å@G ÊÐ&I/*áœë x!u Š Zºnº*Ê¢fŽöÜ&掦 Ü^öN ÷ ˆZ /Y(âš`xäøÑã ý/ÛÉÜe´²¿¦ ’è>õ0§ ¸ö5oËà ...
Web30 mrt. 2024 · Ex9.3,25 If a, b, c and d are in G.P. show that . (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2 We know that a, ar , ar2 , ar3, …. are in G.P. with first term a & … Web30 mrt. 2024 · Example 22 - If a, b, c are in GP and a1/x = b1/y = c1/z. Old search 1. Old search 2.
WebIIT JEE 2002: Let a, b, c be in an AP and a2,b2,c2 be in GP, if a < b < c and a+b+c= (3/2), then the value of a is (A) (1/2 √2) (B) (1/2 √3) (C) (Tardigrade - CET NEET JEE Exam App. Exams; Login; Signup; Tardigrade; Signup; Login; Institution; Exams; Blog; Questions; Tardigrade; Question;
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WebIf a,b,c are in A.P. in a2,b2,c2 are in H.P., then 2247 81 Sequences and Series Report Error A a = b = c B 2b = 3a+c C b2 = 8ac D none of these Solution: Since a,b,c are in A.P. ∴ b −a = c −b ...(1) Since a2,b2,c2 are in H.P. ∴ b21 − a21 = c21 − b21 ⇒ a2b2a2−b2 = b2c2b2−c2 ⇒ a2(a−b)(a+b) = c2(b−c)(b+c) ⇒ a2a+b = c2b+c [Using (1)] psw regulationWeb13 apr. 2024 · Given that are in AP. To prove: are in AP. From given as we know if p , q, r are in AP then 2q= p+r. Now. Which is the result of AP. psw registration processWeb27 jun. 2024 · Explanation: a, b, c are in AP . Therefore a = b-d and c= b+d Given, a+b+c = 3/2 b= 1/2 Hence, a=1/2-d, b=1/2,c=1/2+d Given, a²,b²,c² are in GP. (b²)²= a²c² Solving the above equation, we get, (1/2-d) (1/2+d)= (±1/4) Solve for d, and we get d= 1/√2 (d≥0) a = 1/2 - 1/√2 Advertisement Still have questions? Find more answers Ask your question psw research 2020Web18 sep. 2024 · It is given that a^2 , b^2 and c^2 are in an AP So they have a common difference b^2 - a^2 = c^2-b^2 (b - a)(b + a) = (c - b)(c + b) (b - a) / (b + c) = (c - b) / (b … psw research stationWebTardigrade - CET NEET JEE Exam App. Exams; Login; Signup; Tardigrade; Signup; Login; Institution; Exams; Blog; Questions psw researchWeb18 sep. 2024 · If a2 b2 c2 are in ap then prove that a/b+c b/c+a c/a+b are in ap Advertisement Expert-Verified Answer 89 people found it helpful mailtoaditee It is given that a^2 , b^2 and c^2 are in an AP So they have a common difference b^2 - a^2 = c^2-b^2 (b - a) (b + a) = (c - b) (c + b) (b - a) / (b + c) = (c - b) / (b + a) Let; hort pusteblume münchenWebCorrect option is C) x=1+a+a 2+.....∞= 1−a1 y=1+b+b 2+.....∞= 1−b1 and z=1+c+c 2+.....∞= 1−c1 Since, a,b and c are in AP. ⇒1−a,1−band1−c are also in AP. ⇒ 1−a1, 1−b1 and 1−c1 are in HP. ∴ x, y and z are in HP. Video Explanation Was this … hort rainbow