If 625 Is The Sum Of 5 Consecutive Odd Integers, Then What Is The Largest Of These Integers?

If 625 is the sum of 5 consecutive odd integers, then what is the largest of these integers?

129
131
133
None of these

Explanation

Let the 5 consecutive odd integers be:

x, x+2, x+4, x+6, x+8

Sum = x + (x+2) + (x+4) + (x+6) + (x+8) = 625

5x + 20 = 625

5x = 605

x = 121

So the integers are: 121, 123, 125, 127, 129

Largest = 129

Last verified on 09-06-2026

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1/x3 +1 =(A/x+1 ) + (Bx + C/ x2-x+c) find the value of A is?

1
1/3
-1/3
0

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Explanation

Given:

1/x ^{3} + 1 = A/ x + 1 + B x + C/ x ^{2} - x + 1

Since:

x^{3} + 1 = (x + 1) (x^{2} - x + 1)

Multiplying both sides by: $(x + 1) (x^{2} - x + 1)$ :

1 = A (x^{2} - x + 1) + (B x + C) (x + 1)

Expanding:

1 = (A + B) x^{2} + (- A + B + C) x + (A + C)

Comparing coefficients:

- A + B + C = 0

A + C = 1

A+B=0

B = - A

From $A + C = 1$ :

C = 1 - A

Substitute into the second equation:

- A - A + (1 - A) = 0

1 - 3 A = 0

A =1/ 3

Last updated on 23-07-2026 by NA

Evaluate (1/3)+(1/4)=?

5/12
7/12
4/7
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Simplify (60)² = ? × 72

3600
3528
0.02
50

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a > b
a < b
a = b
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