Question

"For a certain feast, one Brahmin is invited on the first day, and on every succeeding day, one more Brahmin is invited. For another feast, 10 Brahmins are invited every day. In how many days will their numbers be equal, and how many Brahmins were invited?"

Answer 1

After 19 days, the number of Brahmins invited to both feasts will be equal, totalling 190 Brahmins.

Step-by-step explanation:

The student is asking a question that involves sequence and series in mathematics.

Specifically, arithmetic and constant sequences are being compared to find out when the number of Brahmins invited to the two feasts will be equal.

To solve this, let's denote the number of days as 'n'. For the first feast, we can express the total number of Brahmins invited over a period of 'n' days using the arithmetic series formula:

• Sum = n/2 * (first term + last term)
• Sum = n/2 * (1 + n)

For the second feast, the number of Brahmins invited each day is constant at 10, so:

• Total Brahmins for 'n' days = 10n

Now, we set the two expressions equal to find when their numbers are equal:

n/2 * (1 + n) = 10n

Divide both sides by n:

1/2 * (1 + n) = 10

Multiply both sides by 2:

1 + n = 20

Solve for 'n':

n = 19

So, after 19 days, the number of Brahmins invited to both feasts will be equal. To find the total number of Brahmins, we substitute 'n' back into either of the total Brahmins expressions:

Total Brahmins = 10 * 19 = 190 Brahmins.