The sum of first 7 terms of an ap is equal to the sum of its first 13 terms . find the sum of its ist 20 terms
7/2 (2a+6d) = 13/2 (2a+12d)
7a+21d = 13a+78d
6a + 57d = 0
2a + 19d = 0
S20 = 20/2 (2a+19d) = 0
To find the sum of the first 20 terms of an arithmetic progression (AP), we need to determine the common difference (d) first.
Let's assume the first term of the AP is 'a' and the common difference is 'd'.
Given that the sum of the first 7 terms is equal to the sum of the first 13 terms, we can write the equation as:
(7/2) * [2a + (7-1)d] = (13/2) * [2a + (13-1)d]
Simplifying the equation, we get:
7a + 21d = 13a + 78d
Rewriting the equation, we have:
6a = 57d
To find the sum of the first 20 terms, we can use the sum formula for an arithmetic progression:
Sum of the first n terms (Sn) = (n/2) * [2a + (n-1)d]
Let's substitute the values into the formula:
Sum of the first 20 terms = (20/2) * [2a + (20-1)d]
= 10 * [2a + 19d]
= 20a + 190d
From the earlier equation, we know that 6a = 57d.
Let's substitute this value into the sum formula:
Sum of the first 20 terms = 20 * (57d/6) + 190d
= 10 * 57d + 190d
= 570d + 190d
= 760d
Therefore, the sum of the first 20 terms of the arithmetic progression is 760d.
To find the sum of the first 20 terms of the arithmetic progression (AP), we first need to determine the common difference (d) of the AP.
Since the sum of the first 7 terms is equal to the sum of the first 13 terms, we can set up the following equation:
Sum of the first 7 terms = Sum of the first 13 terms
Using the formula for the sum of an AP, we have:
(7/2) * [2a + (7-1)d] = (13/2) * [2a + (13-1)d]
Simplifying this equation, we get:
7 * [2a + 6d] = 13 * [2a + 12d]
14a + 42d = 26a + 156d
12a = 114d
Dividing both sides by 114d, we get:
a/9 = d/19
Now that we have the common difference (d), we can find the sum of the first 20 terms by using the formula:
Sum of the first n terms = (n/2) * [2a + (n-1)d]
Substituting n = 20, a/9 = d/19:
Sum of the first 20 terms = (20/2) * [2a + (20-1)(a/9)(19/9)]
Simplifying this expression will give us the final answer.
Use the formula for the sum of terms
(7/2)(2a + 6d) = (13/2)(2a + 12d)
times 2, then expand
14a + 42d = 26a + 156d
-12a = 114d
a = -19/2 d
sum(20) = 10(2a + 19d)
= 20a + 190d
= 20(-19/2 d) + 190d
= 0