in an arithmetic progression the 13th term is 27 and the 7th term is three times the second term find;the common difference,the first term and the sum of the first ten terms
a+12d = 27
a+6d = 3(a+d)
Now you can find a and d, and then get
S10 = 10/2 (2a+9d)
Let's solve the problem step by step.
Step 1: Find the common difference (d)
We know that the 13th term (a13) is 27 and the 7th term (a7) is three times the second term (a2).
a13 = a1 + (n-1)d
27 = a1 + (13-1)d
a7 = a1 + (7-1)d
a7 = a1 + 6d
Also, a7 = 3(a2)
a1 + 6d = 3(a1 + d)
Simplifying the equations:
27 = a1 + 12d ----------(1)
a1 + 6d = 3a1 + 3d
2a1 = 3d
Step 2: Find the first term (a1)
Using equation (1), substitute 2a1 = 3d
27 = (2a1) + 12d
27 = 3d + 12d
27 = 15d
d = 27/15
d = 9/5
Now, substitute the value of d into 2a1 = 3d
2a1 = 3(9/5)
2a1 = 27/5
a1 = 27/10
So, the first term (a1) is 27/10 and the common difference (d) is 9/5.
Step 3: Find the sum of the first ten terms (S10)
The formula for the sum of the first n terms in an arithmetic progression is:
Sn = (n/2)(2a1 + (n-1)d)
We want to find the sum of the first ten terms (S10), so substitute n = 10, a1 = 27/10, and d = 9/5 into the formula.
S10 = (10/2)(2(27/10) + (10-1)(9/5))
S10 = 5(54/10 + 9(9/5))
S10 = 5(54/10 + 81/5)
S10 = 5(108/20 + 162/20)
S10 = 5(270/20)
S10 = 5(27/2)
S10 = 135/2
Therefore, the sum of the first ten terms (S10) is 135/2.
To find the common difference, we can use the formula for arithmetic progression:
an = a1 + (n-1)d
Where:
an is the nth term of the arithmetic progression,
a1 is the first term,
n is the term number,
and d is the common difference.
Given that an = 27 and n = 13, we can substitute these values into the formula:
27 = a1 + (13-1)d
27 = a1 + 12d
Similarly, for the 7th term, we have:
a7 = a1 + (7-1)d
Given that a7 = 3(a2), we can substitute into the formula:
3(a2) = a1 + 6d
Now, we have a system of equations. We can solve these equations simultaneously using substitution or elimination method.
From equation 2, we get a2 = (a1 + 6d) / 3
Substituting this value into equation 1, we get:
27 = a1 + 12d
27 = a1 + 12[(a1 + 6d) / 3]
27 = a1 + 4(a1 + 6d)
27 = a1 + 4a1 + 24d
Combine like terms:
27 = 5a1 + 24d
Now we can solve this equation for a1 and d.
To find the first term (a1), we can isolate it:
5a1 = 27 - 24d
a1 = (27 - 24d) / 5
Now, let's find the common difference (d):
Substitute the value of a1 into the equation we derived earlier:
27 = 5a1 + 24d
27 = 5((27 - 24d) / 5) + 24d
27 = 27 - 24d + 24d
27 = 27
0 = 0
The equation gives an identity, meaning that d can be any value. The common difference (d) is 0.
So, the common difference is 0, the first term (a1) is (27 - 24d) / 5, where d can be any value, and the sum of the first ten terms can be found using the formula for the sum of an arithmetic progression:
Sn = (n/2)[2a1 + (n-1)d]
Substituting n = 10, a1 = (27 - 24d) / 5, and d = 0 into the formula:
S10 = (10/2)[2((27 - 24d) / 5) + (10-1)(0)]
S10 = 5[(54 - 48d) / 5]
S10 = (54 - 48d)
So, the sum of the first ten terms is (54 - 48d).