Let (a1, b1) and (a2, b2) are the pair of real numbers such that 10, a, b, a b constitute an A.P. Then the value of ((2 a1 a2+b1 b2/10)) is equal to

Question

Let (a1, b1) and (a2, b2) are the pair of real numbers such that 10, a, b, a b constitute an A.P. Then the value of ((2 a1 a2+b1 b2/10)) is equal to (A) 1 (B) 2 (C) 3 (D) 5

Tardigrade Admin · Accepted Answer

Let a =10+ d , b =10+2 d , ab =10+3 d ⇒ (10+d)(10+2 d)=10+3 d ⇒ 2 d2+27 d+90=0 ⇒ d =-6 text or d =(-15/2)

Q. Let $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ are the pair of real numbers such that $10, a, b, ab$ constitute an A.P. Then the value of $(\frac{2 a _{1} a _{2} + b _{1} b _{2}}{10})$ is equal to

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5

Let $a = 10 + d, b = 10 + 2 d, ab = 10 + 3 d$
$\Rightarrow (10 + d) (10 + 2 d) = 10 + 3 d \Rightarrow 2 d^{2} + 27 d + 90 = 0$
$\Rightarrow d = - 6 or d = \frac{- 15}{2}$

Q. Let (a1​,b1​) and (a2​,b2​) are the pair of real numbers such that 10,a,b,ab constitute an A.P. Then the value of (102a1​a2​+b1​b2​​) is equal to

1

2

3

5

Let a=10+d,b=10+2d,ab=10+3d ⇒(10+d)(10+2d)=10+3d⇒2d2+27d+90=0 ⇒d=−6 or d=2−15​

Q. Let $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ are the pair of real numbers such that $10, a, b, ab$ constitute an A.P. Then the value of $(\frac{2 a _{1} a _{2} + b _{1} b _{2}}{10})$ is equal to

Let $a = 10 + d, b = 10 + 2 d, ab = 10 + 3 d$
$\Rightarrow (10 + d) (10 + 2 d) = 10 + 3 d \Rightarrow 2 d^{2} + 27 d + 90 = 0$
$\Rightarrow d = - 6 or d = \frac{- 15}{2}$