The sum of 11 term of an arictiment progratiom is 891 find the 28 and 45 is terms of commion diffrest is 15
11/2 (2a+10*15) = 891
solve for a, and then
a28 = a+27d
a45 = a+44d
To find the 28th and 45th terms of an arithmetic progression when the common difference is 15, we need to find the first term of the sequence.
The formula for the sum of an arithmetic progression is given by:
Sn = (n/2)(2a + (n-1)d)
Where:
Sn = Sum of the first 'n' terms
a = First term of the sequence
d = Common difference between terms
Given that the sum of 11 terms is 891, we can substitute these values into the formula:
891 = (11/2)(2a + (11-1)15)
Simplifying this equation:
891 = (11/2)(2a + 10*15)
891 = (11/2)(2a + 150)
891 = (11/2)(2a + 150)
891 = (11/2)(2a + 150)
891 = 11a + 825
11a = 891 - 825
11a = 66
a = 66/11
a = 6
Now that we have found the first term (a = 6), we can find the nth terms by using the formula:
an = a + (n-1)d
For the 28th term:
a28 = 6 + (28-1)15
a28 = 6 + 27*15
a28 = 6 + 405
a28 = 411
For the 45th term:
a45 = 6 + (45-1)15
a45 = 6 + 44*15
a45 = 6 + 660
a45 = 666
Therefore, the 28th term is 411 and the 45th term is 666 in the arithmetic progression with a common difference of 15.