if sum of 20th term in an AP is equal to sum of 30th term,what is the sum of 50th term of the AP?
This makes no sense. How can one have a sum of one term?
For instance, what is the sum of 21?
Truly
To solve this question, we need to use the formula for the sum of the terms in an arithmetic progression (AP). The sum of an AP is given by:
S = (n/2)(2a + (n-1)d),
where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
Let's denote the sum of the 20th term as S20, and the sum of the 30th term as S30. According to the question, these two sums are equal.
So, we have:
S20 = (20/2)(2a + (20-1)d),
S30 = (30/2)(2a + (30-1)d).
Since S20 = S30, we can equate the two formulas:
(20/2)(2a + 19d) = (30/2)(2a + 29d).
Now, we can simplify this equation by canceling out the common factors and divide both sides by 10:
20(2a + 19d) = 30(2a + 29d).
Simplifying further:
2(2a + 19d) = 3(2a + 29d).
Expanding:
4a + 38d = 6a + 87d.
Rearranging terms:
6a - 4a = 87d - 38d,
2a = 49d.
Dividing both sides by 2:
a = (49d) / 2.
Now we know the relation between the first term 'a' and the common difference 'd' in the AP.
To find the sum of the 50th term, S50, we'll plug in the values of 'a' and 'd' into the formula for the sum of the terms in an AP:
S50 = (50/2)(2 * [(49d) / 2] + (50-1)d).
Simplifying:
S50 = 25(49d + 49d) = 1225d.
Therefore, the sum of the 50th term of the AP is 1225d.
Note: To find the actual value of the sum, you will need the specific values of 'a' and 'd' in the arithmetic progression.