The following questions given below consist of an "Assertion"
≥ (1–2x) (1 – 2y) (1 – 2z) 1/3 Reason : For any three positive number a, b,c, their AM ≥ GM (ii) Assertion: If 
, 
, 
are in AP, then a 1 , b 1 , c 1 are in GP. Reason : If ax 2 + bx + c = 0 and a 1 x 2 + b 1 x + c 1 = 0 have a common root and 
, 
, 
are in AP, then a 1 ,b 1 ,c 1 are in GP. (iii) Assertion : The series for which sum to n terms, S n , is given by S n = 5n 2 + 6n is an A.P. Reason : The sum to n terms of an A.P. having non-zero common difference is a quadratic in n, i.e. an 2 + bn
Text Solution
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Ans.
(i)
Sol. If each of x, y, z is less than 1 then assertion isobviously true.
Also 1 – 2x + 1 – 2y + 1 – 2z = 3–2(x + y + z) = 1
⇒ The sum of the three given number is positive also at most one of x, y, z can be more than 1/2.
⇒ If one of x, y, z is more than or equal to 
, then
their product is less than equal to zero hence still remains true.
Reason is always true but it does not explain assertion.
(ii)
Sol. Since, ax 2 + bx + c = 0 and a 1 x 2 + b 1 x + c 1 = 0 have a common root
(a 1 c–ac 1 ) 2 = 4 (ab 1 –ba 1 ) (bc 1 –b 1 c) ...(i)
If 
, 
, 
are in AP
= 
=k and 
= 2k
On putting these values in Eq. (i), we get c 1 a 1 = b 1 2
(iii)
Sol.
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