The thirteenth term of an arithmetic progrssion os 27 and the seventh term is three timees the second term. tind the first term,common difference and the sum of the first ten terms
term(13) = a+12d
so a+12d = 27 (equation #1)
term(7) = 3term(2)
a+6d = 3(a+d)
a+6d = 3a + 3d
2a - 3d = 0 (equation #2)
It is easy to solve these two equations, (see if you can get a=3, d=-2)
then use your formula for the sum of 10 terms.
Not able to finish
Some how confused about the second equation
To find the first term, common difference, and the sum of the first ten terms in an arithmetic progression, we can use the information provided in the question.
Let's start by finding the common difference (d) using the formula for the nth term of an arithmetic progression:
An = A1 + (n - 1)*d
We are given the thirteenth term (A13 = 27) and the seventh term (A7 = 3*A2).
Substituting the values into the formula, we get:
27 = A1 + (13 - 1)*d
27 = A1 + 12d --> (Equation 1)
A7 = 3*A2
A1 + 6d = 3*(A1 + d)
A1 + 6d = 3A1 + 3d
2A1 = 3d --> (Equation 2)
Using the two equations above, we can solve for A1 and d simultaneously.
From (Equation 2), we can express A1 in terms of d:
A1 = (3/2)d
Substituting A1 into (Equation 1):
27 = (3/2)d + 12d
27 = (27/2)d
d = 2
Now that we have found the common difference (d = 2), we can find the first term (A1) by substituting d into Equation 2:
A1 = (3/2)*2
A1 = 3
So, the first term (A1) is 3, and the common difference (d) is 2.
Lastly, we can find the sum of the first ten terms (S10) using the formula for the sum of an arithmetic progression:
Sn = (n/2)(2A1 + (n - 1)d)
Substituting the values, we get:
S10 = (10/2)(2*3 + (10 - 1)*2)
S10 = 5(6 + 18)
S10 = 5 * 24
S10 = 120
Therefore, the sum of the first ten terms (S10) is 120.