Question

find 4 numbers in Ap such that their sum is 26 and product is 880.

Answer 1

Let the four numbers be (a-3d), (a-d), (a+d) & (a+3d).

Now A.T.Q

1. Their sum

(a-3d)+(a-d)+(a+d)+(a+3d)=26

4a = 26

a=6.5

2.Their product

(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=880

4a²+20d²=880

a² + 5d² = 220

Putting the value of a

42.25 + 5d² = 54

5d² = 11.25

d² = 2.35

d=±√2.35

hope it is helpful

Hence, the numbers are

Answer 2

Concept :

It is also known as Arithmetic Sequence. Arithmetic progression (AP) is a sequence of numbers in order, where the difference between any two consecutive numbers is a constant value.

nth term of an AP = a + (n-1) d.

Arithmetic Mean = Sum of all terms in the AP / Number of terms in the AP.

Sum of 'n' terms of an AP = 0.5 n (first term + last term) = 0.5 n [ 2a + (n-1) d ]

Given :

4 numbers in Ap such that their sum is 26 and product is 880.

Find :

Find 4 numbers in Ap such that their sum is 26 and product is 880.

Solution :

($a$-$3d)+(a-d)+(a+d)+(a+3d)=26$

$4a = 26$

$a=6.5$

2.Their product

$(a-3d)²+(a-d)²+(a+d)²+(a+3d)²=880$

$4a²+20d²=880$

$a² + 5d² = 220$

Putting the value of a

$42.25 + 5d² = 54$

$5d² = 11.25$

$d² = 2.35$

$d=±√2.35$

Hence the number is founded

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