If a , b , c are in A . P . , then show that 10 a x + 10 , 10 b x + 10 , 10 c x + 10 , x ≠ 0 , are in G . P .

Given that, a , b , c are in AP . If x , y , z are three consecutive terms in AP then 2 y = x + z . ⇒ 2 b = a + c ...(i) Verifying 10 a x + 10 , 10 b x + 10 , 10 c x + 10 are in GP or not. If p , q , s are three consecutive terms in GP then q 2 = p s . ⇒ 10 b x + 10 2 = 10 a x + 10 10 c x + 10 ⇒ 10 2 ( b x + 10 ) = 10 a x + 10 + c x + 10 ⇒ 10 2 b x + 20 = 10 a x + c x + 20 As bases are equal, indices must be equal. ⇒ 2 b x + 20 = a x + c x + 20 ⇒ 2 b x = a x + c x ⇒ 2 b x = x ( a + c ) ⇒ 2 b = a + c is true. [ from (i) ] ∴ 10 a x + 10 , 10 b x + 10 , 10 c x + 10 are in GP . Hence, proved.

If k , 2 k + 2 , 3 k + 3 , … are in G . P . , then find the common ratio of the G . P .

Given that, k , 2 k + 2 , 3 k + 3 , ⋯ are in GP . If p , q , s are three consecutive terms in GP then q 2 = p s . ⇒ ( 2 k + 2 ) 2 = k ( 3 k + 3 ) ⇒ 4 k 2 + 8 k + 4 = 3 k 2 + 3 k ⇒ k 2 + 5 k + 4 = 0 ⇒ k 2 + 4 k + k + 4 = 0 ⇒ ( k + 4 ) ( k + 1 ) = 0 ⇒ k = - 4 , - 1 At k = - 1 , 3 k + 3 = 0 . In a GP no term can be zero. ∴ k = - 4 ∴ GP is - 4 , 2 ( - 4 ) + 2 , 3 ( - 4 ) + 3 , ⋯ i.e., - 4 , - 6 , - 9 , ⋯ Common ratio, r = a 2 a 1 = - 6 - 4 ∴ Common ratio, r = 3 2 .