Question

the sum of three consecutive terms in AP is 54 and the product of two extremes is 275 find the numbers​

Answer 1

Step-by-step explanation:

let the three terms be

a-d,a,a+d

Now, sum = 54

a-d+a+a+d = 54

3a= 54

a=18

Also, (a-d)(a-d)=275

(a×a)-(d×d)=275

18^2 - d^2=275

324 -d^2 = 275

d^2= 49

d=7,-7

numbers= 11,18,25

or

25,18,11

Answer 2

Numbers are 11,18,25 .

Step-by-step explanation:

Given : The sum of three consecutive terms in AP is 54 and the product of two extremes is 275.

To find : The numbers​ ?

Solution :

Let the three consecutive terms in A.P. are a-d, a, a+d .

The sum of three consecutive terms in AP is 54.

i.e. $a-d+a+a+d=54$

$3a=54$

$a=\frac{54}{3}$

$a=18$

The first term is a=18.

The product of two extremes is 275.

i.e. $(a-d)(a-d)=275$

$a^2-d^2=275$

$(18)^2-d^2=275$

$-d^2=-49$

$d=\sqrt{49}$

$d=\pm7$            

The common difference are 7 or -7.

The A.P form is when a=18 and d=7

a-d=18-7=11

a=18

a+d=18+7=25

Numbers are 11,18,25 .

The A.P form is when a=18 and d=-7

a-d=18-(-7)=25

a=18

a+d=18-7=11

Numbers are 25,18,11 .

#Learn more

In an ap the first term is -5 and last term is 45.if the sum of all numbers in the ap is 120 then how many terms are there and what is the commom differnce

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