Math, asked by vivekraj4407, 1 year ago

Find 4 numbers in ap whose sum is 20 and the sum of whose square is 120
In and why we take a-3d,a-d,a+d,a+3d

Answers

Answered by mysticd
2

Solution :

Let a-3d,a-d,a+d,a+3d are

4 terms in A.P

According to the problem given,

a-3d+a-d+a+d+a+3d= 20

=> 4a = 20

=> a = 20/4

=> a = 5

(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=120

=> 2[a² + (3d)²]+2[a²+d²] = 120

=> a² + 9d² + a² + d² = 60

=> 2a² + 10d² = 60

=> 2×5² + 10d² = 60

=> 50 + 10d² = 60

=> 10d² = 60 - 50

=> d² = 10/10

=> d² = 1

=> d = ± 1

Now ,

Required 4 numbers of A.P :

i ) if a = 5 , d = 1

(5-3),(5-1),(5+1),(5+3)

i.e , 2 ,4 , 6, 8

ii ) if a = 5 , d = -1

4 terms are

8,6,4,2

***********

We take a-3d,a-d,a+d,a+3d

to get first term ( a ) value

easily .

Otherwise it will be lengthy

procedure to solve the two

equations to get a , d values .

••••

Answered by amitnrw
2

Answer:

2 , 4 , 6 , 8

Step-by-step explanation:

Let say four numbers are

a - 3d  , a -d , a + d , a + 2d

Reason why we take a-3d,a-d,a+d,a+3d

So that we sum all these numbers we get only on variable and we can find value of a

But it is not necessary to take number like this

we can take a , a+d , a+2d , a+3d also. check my alternate solution after its completion

Sum of a-3d,a-d,a+d,a+3d = 20

a -3d + a -d + a + d + a + 3d = 20

=> 4a = 20

=> a = 5

(a-3d)² + (a-d)² + (a+d)² + (a+3d)² = 120

=> a² + 9d² - 6ad + a² + d² - 2ad +  a² + d² + 2ad + a² + 9d² + 6ad = 120

=> 4a² + 20d² = 120

=> a² + 5d² = 30

a = 5

=> 5² + 5d² = 30

=> 25 + 5d² = 30

=> 5d² = 30 -25

=> 5d² = 5

=> d² = 1

=> d = ±1

So numbers are

2 , 4 , 6 , 8    or  8 , 6 , 4 , 2

____________________________

here alternate solution

Let say 4 numbers are

a , a+d , a+2d , a + 3d

=> 4a + 6d = 20

=> 2a + 3d = 10   - eq 1

a² + (a+d)² + (a+2d)² + (a+3d)² = 120

=> a² + a² + d² + 2ad + a² + 4d² + 4ad + a² + 9d² + 6ad = 120

=> 4a² + 14d² + 12ad = 120   Eq2

Squaring Eq 1

(2a + 3d)² = 10²

=> 4a² + 9d² + 12ad = 100   Eq3

Eq 2 - Eq 3

=> 5d² = 20

=> d² =4

=> d = ±2

Now putting values of d in eq 1  ,  2a + 3d = 10

Case 1 d = 2

2a + 6 = 10

=> 2a = 4

=> a = 2

numbers are  2 , 4 , 6 & 8

Case 2 d = -2

2a - 6 = 10

=> 2a = 16

=> a = 8

Numbers are 8 , 6 , 4 & 2

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