Answer : To prove: x2, b2, y2 are in AP. Given: a, b, c are in AP, and a, x, b an b, y, c are in GP Proof: As, a,b,c are in AP ⇒ 2b = a + c … (i) As, a,x,b are in GP ⇒ x2 = ab … (ii) As, b,y,c are...
Answer : To prove: a, (a – b) and (d – c) are in GP. Given: a, b, c are in AP, and a, b, d are in GP Proof: As a,b,d are in GP then b2 = ad … (i) As a, b, c are in AP 2b = (a + c) … (ii) Considering...
Answer : To prove: p, q, r are in GP Given: (p2 + q2), (pq + qr), (q2 + r2) are in GP Formula used: When a,b,c are in GP, b2 = ac Proof: When (p2 + q2), (pq + qr), (q2 + r2) are in GP, (pq + qr)2 =...
Answer : To prove: (a2 – b2), (b2 – c2), (c2 – d2) are in GP. Given: a, b, c are in GP From the above, we can have the following conclusion Formula used: When a,b,c are in GP, b2 = ac Proof: When...
Answer : To prove: (a2 + b2), (ab + bc), (b2 + c2) are in GP Given: a, b, c are in GP Formula used: When a,b,c are in GP, b2 = ac Proof: When a,b,c are in GP, b2 = ac … (i) Considering (a2 + b2),...
Answer : To prove: a3, b3, c3 are in GP Given: a, b, c are in GP Proof: As a, b, c are in GP ⇒ b2 = ac Cubing both sides = common ratio = r From the above equation, we can say that a3, b3, c3 are in...
Answer : To prove: a2, b2, c2 are in GP Given: a, b, c are in GP Proof: As a, b, c are in GP ⇒ b2 = ac … (i) Considering b2, c2 = common ratio = r ⇒ [From eqn. (i)] ⇒ = r Considering a2,...
Hence Proved (iii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2 To prove: (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2 Given: a, b, c, d are in GP Proof: When a,b,c,d are in GP thenFrom...
Answer : To prove: (a + b + c)2 = 3(ab + bc + ca). Given: (a – b), (b – c), (c – a) are in GP Formula used: When a,b,c are in GP, b2 = ac As, (a – b), (b – c), (c – a) are in GP ⇒ (b – c)2 = (a – b)...
Answer : To prove: Given: a, b, c are in GP Formula used: When a,b,c are in GP, b2 = ac a, b, c are in GP, ⇒ b2 = ac … (i) Taking LHS Substituting the value b2 = ac from eqn. (i) Substituting the...
Answer : To find: Three numbers Given: Three numbers are in G.P. Their sum is 56 Formula used: When a,b,c are in GP, b2 = ac Let the three numbers in GP be a, ar, ar2 According to condition :- a +...
Answer : To find: Three numbers Given: Three numbers are in A.P. Their sum is 21 Formula used: When a,b,c are in GP, b2 = ac Let the numbers be a - d, a, a + d According to first condition a + d + a...
Answer : To find: The numbers Given: Three numbers are in A.P. Their sum is 15 Formula used: When a,b,c are in GP, b2 = ac Let the numbers be a - d, a, a + d According to first condition a + d + a...
Answer : To find: Value of k Given: k + 12, k – 6 and 3 are in GP Formula used: (i) when a,b,c are in GP b2 = ac As, k + 12, k – 6 and 3 are in GP ⇒ (k – 6)2 = (k + 12) (3) ⇒ k2 – 12k + 36 = 3k + 36...
Answer : To prove: log an, log bn, log cn are in AP. Given: a, b, c are in GP Formula used: (i) log ab = log a + log b As a, b, c are in GP ⇒ b2 = ac Taking power n on both sides ⇒ b2n = (ac)n...
To prove: pth, qth and rth terms of any GP are in GP. Given: (i) p, q and r are in AP The formula used: (i) General term of GP, As p, q, r are in A.P.