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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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Hi,

1850.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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Hi,

1851.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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Hi,

1852.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**cheetahflycello****Novice**- Registered: 2023-04-19
- Posts: 6

For the arithmetic progression problem, let's call the three terms a-d, a, and a+d (where 'd' is the common difference). We know their sum is -6, so we have: (a-d) + a + (a+d) = -6. Simplifying gives us 3a = -6, and therefore, a = -2. Now, let's find the common difference 'd': (-2) + a + (-2+d) = -6. Solving this gives us d = -2. Now we can find the three terms: -2-(-2) = -4, -2, -2+(-2) = -4. So the three terms are -4, -2, and -4.

For the cylindrical jar and leadshots problem, first, let's find the volume of the jar when it's half-full of water. The volume of a cylinder is V = π * r^2 * h, where 'r' is the radius and 'h' is the depth. The radius is half of the diameter, so r = 14 cm / 2 = 7 cm. When half-full, the water level is 20 cm / 2 = 10 cm. So, the volume of water in the jar is V_water = π * 7^2 * 10 = 490π cm³.

After dropping the 300 leadshots, the water level rises by 2.8 cm, so the new depth is 20 cm + 2.8 cm = 22.8 cm. Let's call the diameter of each leadshot 'd'. The volume of one leadshot is V_leadshot = (π * d^3) / 6. The total volume added by the 300 leadshots is 300 * V_leadshot. Since the total volume added equals the increase in water volume (490π cm³), we can set up the equation: 300 * (π * d^3) / 6 = 490π.

Now, we can solve for 'd': 300 * d^3 = 2940. Dividing both sides by 300 gives d^3 = 9.8. Finally, taking the cube root of both sides, we get d ≈ 2.15 cm (rounded to two decimal places).

So, the diameter of each leadshot is approximately 2.15 centimeters.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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Hi cheetahflycello,

Where are the problems?

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1843.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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Hi,

1844.

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**SHANTAMURTHY HIREMATH****Novice**- Registered: 2023-07-28
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Jai Ganesh wrote:

1. The sum of three terms of an Arithmetic Progression is -6 and their product is 90. Find the three terms.

2. A cylindrical jar of diameter 14 centimeters and depth 20 centimeters is half-full of water. 300 leadshots of same size are dropped into the jar and the water level raises by 2.8 centimeters. Find the diameter of each leadshot.

Solution: let the terms of an ap be a-d, a and a+d

sum of these will be 3a=-6, then we get the value of a as -2.

as per the second statement. -8+2 dsquare = 90

then we will have d squared= 49

then value of d is 7

hence the terms are -9, -2, 5

thank you for posting this question..

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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Now go to compute 1844 and solve!

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1845.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1846.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1847.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1848.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1849.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1850.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1851.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1852.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1853.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1854.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1855.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1856.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1857.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1858.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1859.

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
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1860.

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