The sum of 5th term and 24th term is 50
Find the sum of first term and 28th term.
Also find sum of 28th term of this sequence?
T5 + T24 = 50
a+4d + a+23d = 50
a + a+4d+23d = 50
a + a+27d = 50
T1 + T28 = 50
I assume you are dealing with an AS
a+4d + a+23d = 50
2a + 27d = 50
sum of first and 28th
= a + a+27d
= 2a + 27d
= 50
28th term = a + 27d
= 50 - a
the first term of a.p is 50 and the fifth term is 25 .find the sum of the first 10th term
ade mistakenly approximate 0.005642 to 2s.f instead of 2d.p. what the percentage error correct to 2 d.p
To find the sum of the first term and 28th term, we can set up an equation using the information given about the 5th term and 24th term:
Let the first term be represented by a.
The 5th term would then be a + 4d, where d represents the common difference between the terms.
The 24th term would be a + 23d.
According to the given information, the sum of the 5th term and 24th term is 50:
(a + 4d) + (a + 23d) = 50
Simplifying this equation, we get:
2a + 27d = 50 ––> Equation 1
Now, to find the sum of the first term and 28th term, we can use the equation we just found. The 28th term would be a + 27d, so the sum of the first term and 28th term can be represented as:
a + (a + 27d) = 2a + 27d
Therefore, the sum of the first term and 28th term is equivalent to 2a + 27d.
Now, to find the sum of the 28th term, we can use another method. Since we know the 28th term is a + 27d, we can find the sum of all terms up to the 28th term using the formula for the sum of an arithmetic series:
Sum = (n/2)(2a + (n-1)d)
Here, n represents the number of terms we want to sum up, which in this case is 28.
Therefore, the sum of the 28th term would be:
(28/2)(2a + 27d)
Simplifying further, we get:
14(2a + 27d) = 28a + 378d
So, the sum of the 28th term in the sequence is equivalent to 28a + 378d.