If 1 A 1 B 1 C Are In A P Then 1 A 1 B 1 C 1 B 1 C 1 A Is

If 1 A 1 A 2b 1 C 1 C 2b 0 And A B C Are Not In A P Then If 1 a, 1 b, 1 c are in a.p., prove that: (i) (b c) a, (c a) b, (a b) c are in a.p. (ii) a (b c), b (c a), c (a b) are in a.p. challenge your friends with exciting quiz games – click to play now! hence, the given terms are in ap. hence, given terms are in ap. Perhaps the best way to approach this question is to assume ap, gp and hp and see whether the identity holds; for, indeed, should the solution be d), you will be performing algebraic manipulations on the identity to no avail unless you can put it in a form which encompasses all 3 possibilities simultaneously.

If 1 A 1 B 1 C Are In A P Prove That I B C A C A B A ⇒ b2 – a2 = c2 – b2. ⇒ a2 , b2 , c2 are in a.p. Q. if a1, b1, c1 are in a. p., then (a1 b1 − c1)(b1 c1 − a1) is equal to. Misc 16 if a (1 (𝑏 ) " " 1 𝑐), b (1 𝑐 " " 1 𝑎), c (1 𝑎 " " 1 𝑏) are in ap., prove that a, b, c are in ap given that a (1 (𝑏 ) " " 1 𝑐), b (1 𝑐 " " 1 𝑎), c (1 𝑎 " " 1 𝑏) are in ap. if a (1 (𝑏 ) " " 1 𝑐), b (1 𝑐 " " 1 𝑎), c (1 𝑎 " " 1 𝑏) are in ap adding 1 to each term a (. Solution: it is given that a (1 b 1 c), b (1 c 1 a), c (1 a 1 b) are in a.p. therefore, b (1 c 1 a) a (1 b 1 c) = c (1 a 1 b) b (1 c 1 a) ⇒ b (a c) ac a (b c) bc = c (a b) ab b (a c) ac. ⇒ (b 2 a b 2 c a 2 b a 2 c) abc = (c 2 a c 2 b b 2 a b 2 c) abc.

If 1 A 1 B 1 C Are In A P Prove That 1 B C A C A B A B C Are Misc 16 if a (1 (𝑏 ) " " 1 𝑐), b (1 𝑐 " " 1 𝑎), c (1 𝑎 " " 1 𝑏) are in ap., prove that a, b, c are in ap given that a (1 (𝑏 ) " " 1 𝑐), b (1 𝑐 " " 1 𝑎), c (1 𝑎 " " 1 𝑏) are in ap. if a (1 (𝑏 ) " " 1 𝑐), b (1 𝑐 " " 1 𝑎), c (1 𝑎 " " 1 𝑏) are in ap adding 1 to each term a (. Solution: it is given that a (1 b 1 c), b (1 c 1 a), c (1 a 1 b) are in a.p. therefore, b (1 c 1 a) a (1 b 1 c) = c (1 a 1 b) b (1 c 1 a) ⇒ b (a c) ac a (b c) bc = c (a b) ab b (a c) ac. ⇒ (b 2 a b 2 c a 2 b a 2 c) abc = (c 2 a c 2 b b 2 a b 2 c) abc. Since, (b c) a, (c a) b, (a b) c are in ap a (b c) = c (a b) lhs = rhs hence, the given terms are in a.p (ii) bc, ca, ab are in ap if bc, ca, ab are in ap then, ca – bc = ab – ca c (a b) = a (b c) if 1 a, 1 b, 1 c are in ap then, 1 b – 1 a = 1 c – 1 b c (a b) = a (b c) hence, the given terms are in ap. Given :– terms 1 a , 1 b , 1 c are in a.p. to prove :– (b c) a , (c a) b , (a b) c are in a.p. solution :– • according to the question – => 1 a , 1 b , 1 c are in a.p. • now multiply with (a b c) — • now , we should write this as – • now subtract 1 from all terms – (hence proved). If the sum of m terms of an a.p. is equal to the sum of either the next n terms or the next p terms, then prove that (m n) (1 m 1 p) = (m p) (1 m 1 n) a man accepts a position with an initial salary of rs 5200 per month. Hence, option (b) is correct. q. if α, β, γ are the roots of x3 ax2 b=0, b≠0 then the determinant Δ, where. q. among 1 12 , 1 15 and 1 17 , is the greatest fraction. Δ=0 reason: Δ can be written as product of two determinants. q.