Web6 aug. 2024 · If a,b,c are in GP and a + x , b + x, c + x are in HP. Algebra. 2 Answers WebSolution The correct option is C H. P. Step 1. Finding the value of a, b, c: Given, ( b + c - a) a, ( c + a - b) b, ( a + b - c) c are in A P Add 2 to all terms. ⇒ b + c - a a + 2, c + a - b b + 2, a + b - c c + 2 are in A P. ⇒ b + c - a + 2 a a, c + a + b + 2 b b, a + b - c + 2 c c are in A P. ⇒ a + b + c a, a + b + c b, a + b + c c are in A P.
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Web30 jul. 2024 · If a, b, c are in AP, x is the GM between a and b; y is the GM between b and c; then show that b^2 is the AM. If a, b, c are in AP, x is the GM between a and b; y is … Web30 mrt. 2024 · If a, b, c, are in A.P., then the determinant 8 (𝑥+2&𝑥+3&𝑥+2𝑎@𝑥+3&𝑥+4&𝑥+2𝑏@𝑥+4&𝑥+5&𝑥+2𝑐) is A. 0 B. 1 C. x D. 2x Since a, b & c are in A.P …
WebIf the roots of the equation (b - c) x2 + (c - a)x + (a - b) = 0 be equal, then prove that a, b, c are in arithmetic progression. Solution Given, (b-c) x2 + (c - a)x + (a - b) = 0 Comparing given equation with general form Ax3+Bx+C= 0 We get, A=(b−c),B=(c−a),D= (a−b) If roots are equal then discriminant is equal to zero. That is, D= B2−4AC =0 Web6 aug. 2024 · Explanation: As a,b and c are in GP, we have b2 = ac. Further as a,b,c are distinct numbers, a ≠ b ≠ c ≠ a. If a +x, b +x and c +x are in HP, then. 1 a +x, 1 b +x and …
Web2 mei 2024 · Three numbers a, b, c are in GP. and ax = by = c^z, then prove that x, z will be in H.P. asked May 2, 2024 in Sequence, Progression, and Series by PritiKumari ( 49.2k points) sequence Web20 apr. 2024 · Step-by-step explanation: Since , a, b ,c are in AP; So, equating the common difference of the given AP; b-a = c-b b = (a+c)/2........... (1) Now, since; b-a , c-b and a …
Web16 jul. 2024 · If a, b, c are in G.P. and x, y are AM’s between a, b and b, c respectively, then A.1/x + 1/y = 2 B. 1/x + 1/y = 1/2 ← Prev Question Next Question → 0 votes 94 views asked Jul 16, 2024 in Geometric Progressions by kavitaKumari (13.5k points) If a, b, c are in G.P. and x, y are AM’s between a, b and b, c respectively, then
Web9 apr. 2024 · a, b, c are in AP b = a + c 2 Also, A. M > G. M a + b 2 > a b sequences-and-series algebra-precalculus arithmetic-progressions Share Cite Follow asked Apr 9, 2024 at 2:10 pi-π 7,376 6 76 158 Note that a, b, c can be negative. I would not use AM/GM, just take b = ( a + c) / 2 as you have done and write out equivalences for the inequality b 2 > … dennis most south dakotaWebCorrect 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 ( … dennis moses attorneyWeba, b and c are in AP. ⇒ 2b = a + c a, x and b are in GP. ⇒ x 2 = ab b, y and c are in GP. ⇒ y 2 = bc now. x 2 + y 2 = ab + bc = b(a+ c) = b x 2b = 2b 2 ⇒ x 2, b 2 and y 2 are in A.P. dennis mouland advanced cadastral academyWeb22 mrt. 2024 · Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. dennis most and the instigatorsWeb9 apr. 2024 · Note that a, b, c can be negative. I would not use AM/GM, just take b = ( a + c) / 2 as you have done and write out equivalences for the inequality b 2 > a c. [This of … ffmhcWeb29 jul. 2024 · If a, b, c are in AP, and a, x, b and b, y, c are in GP then show that x^2 , b^2 , y^2are in AP. asked Jul 29, 2024 in Geometric Progressions by KumarArun (14.8k points) geometric progressions; class-11; Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to ... ffmgf priceWeb30 jul. 2024 · Given: (i) a, b, c are in AP (ii) x is the GM between a and b (iii) y is the GM between b and c Formula used: (i) Arithmetic mean between a and b = a+b 2 a + b 2 (ii) Geometric mean between a and b = √ab a b As a, b, c are in A.P. ⇒ 2b = a + c … (i) As x is the GM between a and b ⇒ x = (√ab) ( a b) ⇒ x2 = ab … (ii) As y is the GM between b … ffm goetheturm