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Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2

Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2
Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2

Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2 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 & common ratio r given a, b, c, d are in g.p. so, a = a b = ar c = ar2 d = ar3 we need to sho. If a = 1 b b2 b3 to ∞, then write b in terms of a. in a g.p. of even number of terms, the sum of all terms is five times the sum of the odd terms.

Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2
Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2

Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2 Ncert solutions class 11 maths chapter 9 exercise 9.3 question 25. if a, b, c an d are in g.p. show that: (a 2 b 2 c 2) (b 2 c 2 d 2) = (ab bc cd) 2. summary: given that a, b, c, d in g.p we showed that a 2 b 2 c 2) (b 2 c 2 d 2) = (ab bc cd) 2. To show that (ab bc cd)2 = (a2 b2 c2)(b2 c2 d2) given that a,b,c,d are in geometric progression (g.p.), we can follow these steps: expanding both sides will show that they are equal. if a, b, c and d are in g.p. show that (a2 b2 c2)(b2 c2 d2) = (ab bc cd)2. if a,b,c,d be in g.p. show that (b− c)2 (c−a)2 (d −b)2 = (a − d)2. Ex 9.3 class 11 q25 (a2 b2 c2) (b2 c2 d2)= (ab bc cd)2 if a b c d are in gp then (a2 b2 c2) (b2 c2 d2) more. If a, b, c, d are in g.p., then ( (a2 b2 c2) (b2 c2 d2) (ab bc cd)2)= (a) 1 (b) 2 (c) 3 (d) 5. check answer and solution for above question from mat.

Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2
Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2

Ex 8 2 25 If A B C D Are In Gp Show A2 B2 C2 Ex 9.3 class 11 q25 (a2 b2 c2) (b2 c2 d2)= (ab bc cd)2 if a b c d are in gp then (a2 b2 c2) (b2 c2 d2) more. If a, b, c, d are in g.p., then ( (a2 b2 c2) (b2 c2 d2) (ab bc cd)2)= (a) 1 (b) 2 (c) 3 (d) 5. check answer and solution for above question from mat. If $a, b, c, d$ are in g.p., show that (i) $a^ {2} b^ {2}, b^ {2} c^ {2}, c^ {2} d^ {2}$ are n g.p. (ii) $\frac {1} {a^ {2} b^ {2}}, \frac {1} {b^ {2} c^ {2}}, \frac {1} {c^ {2} d^ {2}}$ are in g.p. A, b, c, d are in g.p. therefore, bc = ad … (1) b2 = ac … (2) c2 = bd … (3) it has to be proved that, (a2 b2 c2) (b2 c2 d2) = (ab bc – cd)2. r.h.s. = (ab bc cd)2. = (ab ad cd)2 [using (1)] = [ab d (a c)]2. = a2b2 2abd (a c) d2 (a c)2. = a2b2 2a2bd 2acbd d2 (a2 2ac c2). If a, b, c and d are in g.p. show that (a^2 b^2 c^2) (b^2 c^2 d^2) = (ab bc cd)^2 . ← prev question next question → 4 votes 29.4k views. If a b x a b x = b c x b c x = c d x c d x (x ≠ 0) then show that a, b, c and d are in g.p. if a and b are the roots of are roots of x 2 – 3x p = 0 , and c, d are roots of x 2 – 12x q = 0, where a, b, c, d, form a g.p. prove that (q p): (q – p) = 17 : 15.

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