the sum of 11 term of an A.p is 891 find the 28 and 45 terms if the common difference is 15
the sum of 11 term of an A.p is 891 find the 28 and 45 terms if the common difference is 15
Yeah -- we got it the first time
11/2 (2a+10*15) = 891
a = 6
a28 = a + 27d
a45 = a + 44d
So just plug and chug
To find the 28th and 45th terms of an arithmetic progression (A.P.), we need to use the given information of the sum of 11 terms and the common difference.
Let's start by finding the first term (a) using the given formula for the sum of n terms in an A.P.:
Sn = (n/2)[2a + (n-1)d]
Where:
Sn = Sum of n terms
a = First term
n = Number of terms
d = Common difference
We know that the sum of 11 terms (Sn) is 891:
891 = (11/2)[2a + (11-1)(15)]
891 = (11/2)[2a + 10(15)]
891 = (11/2)[2a + 150]
891 = 11a + 825
891 - 825 = 11a
66 = 11a
a = 66/11
a = 6
Now we know that the first term (a) is 6.
To find the 28th term, we use the formula for the nth term in an A.P.:
An = a + (n-1)d
Substituting the values:
A28 = 6 + (28-1)(15)
A28 = 6 + 27(15)
A28 = 6 + 405
A28 = 411
Therefore, the 28th term of the A.P. is 411.
Similarly, to find the 45th term, we use the same formula:
A45 = 6 + (45-1)(15)
A45 = 6 + 44(15)
A45 = 6 + 660
A45 = 666
Therefore, the 45th term of the A.P. is 666.
To find the 28th and 45th terms of an arithmetic progression (A.P.) when the common difference is 15, we first need to calculate the first term (a) and the common difference (d).
We are given that the sum of 11 terms of the A.P. is 891. Given the formula for the sum of an A.P., Sn = (n/2)(2a + (n-1)d), we can substitute the given values:
891 = (11/2)(2a + 10d)
To simplify the equation, we can multiply both sides by 2 to get rid of the fraction:
1782 = 11(2a + 10d)
Next, we can distribute the 11 to simplify further:
1782 = 22a + 110d
Now we have a system of equations with two variables:
Equation 1: 1782 = 22a + 110d
Equation 2: d = 15 (given)
We can substitute the value of d into Equation 1:
1782 = 22a + 110(15)
1782 = 22a + 1650
Now, let's isolate the variable a:
22a = 1782 - 1650
22a = 132
a = 132 / 22
a = 6
So, the first term (a) of the A.P. is 6.
Now, let's calculate the 28th term (T28) and 45th term (T45) using the formula for the nth term of an A.P., which is Tn = a + (n-1)d.
T28 = 6 + (28 - 1)15
T28 = 6 + 27(15)
T28 = 6 + 405
T28 = 411
T45 = 6 + (45 - 1)15
T45 = 6 + 44(15)
T45 = 6 + 660
T45 = 666
So, the 28th term of the A.P. is 411 and the 45th term is 666.