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The Sum of three numbers in AP is 75, and product of extremit

Question

The Sum of three numbers in AP is $75$ , and product of extremities is $609$ . The numbrs and AM of first and second number is

{21, 25, 29}

, AM

= 23

{13, 17, 21}

, AM

= 22

{21, 25, 29}

, AM

= 25

{21, 22, 29}

, AM

= 23

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The Sum of three numbers in AP is

75

, and product of extremities is

609

. The numbrs and AM of first and second number is

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

Let the numbers in A.P. be (a-d), a, (a+d)

∴ (a - d) + a + (a + d) = 75

\Rightarrow a = 25

Also,

(a - d) (a + d) = 609

\Rightarrow a^{2} - d^{2} = 609

\Rightarrow 2 5^{2} - d^{2} = 609

∴ d = \pm 4

The numbers are {21,25,29} or {29,25,21}

According to option, we take {21,25,29}

A.M. of first and second number

= \frac{21 + 25}{2} = 23

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Question Text	The Sum of three numbers in AP is $75$ , and product of extremities is $609$ . The numbrs and AM of first and second number is
Answer Type	Text solution:1
Upvotes	150