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
To solve this problem, let's break it down step by step.
Step 1: Finding the common difference (d)
We know that the thirteenth term of the arithmetic progression is 27. The general formula to find any term in an arithmetic progression is:
tn = a + (n - 1)d
where:
tn = nth term
a = first term
n = term number
d = common difference
Using the given information, we can substitute tn = 27 and n = 13 into the formula:
27 = a + (13 - 1)d
27 = a + 12d
Step 2: Finding the relation between the second term (t2) and the seventh term (t7)
The problem states that the seventh term is three times the second term. We can express this relation as:
t7 = 3t2
Substituting the terms using the general formula:
a + (7 - 1)d = 3(a + (2 - 1)d)
a + 6d = 3a + 2d
4d = 2a
2d = a
Step 3: Finding the first term (a)
Using the relation we found, we can substitute 2d for a in the equation from Step 1:
27 = 2d + 12d
27 = 14d
Divide both sides by 14:
d = 27/14
Step 4: Finding the first term (a)
Now that we know the value of d, we can substitute it back into the equation we found in Step 3:
2d = a
2(27/14) = a
a = 54/14
a ≈ 3.857
So the first term (a) of the arithmetic progression is approximately 3.857, and the common difference (d) is 27/14.
Step 5: Finding the sum of the first ten terms (S10)
To find the sum of the first ten terms, we can use the formula for the sum of an arithmetic progression:
Sn = (n/2) * (2a + (n - 1)d)
Substituting the values we found:
n = 10, a ≈ 3.857, d = 27/14
S10 = (10/2) * [2(3.857) + (10 - 1)(27/14)]
S10 = 5 * [7.714 + 9(27/14)]
S10 = 5 * [7.714 + (243/14)]
S10 = 5 * (7.714 + 17.357)
S10 = 5 * 25.071
S10 = 125.355
Therefore, the sum of the first ten terms (S10) is approximately 125.355.