The 3rd term of an ap is 15 and the 15th term is 21 find the common difference first term and the sum of the first ten te
Common difference = 6
First term = 3
Sum of first ten terms = 135
To find the common difference, first consider the given information:
The 3rd term is 15, and the 15th term is 21.
We can use these two terms to find the common difference.
Step 1: Find the difference between the 15th term and the 3rd term.
21 - 15 = 6
Step 2: Find the difference in position between the 15th term and the 3rd term.
15 - 3 = 12
Step 3: Divide the difference in values (Step 1) by the difference in positions (Step 2).
6 ÷ 12 = 0.5
Therefore, the common difference of this arithmetic progression (AP) is 0.5.
To find the first term of the AP, we can use the formula:
first term (a) = any given term (n) - (common difference (d) × (position of the given term - 1))
Using the formula with the given values:
a = 15 - (0.5 × (3 - 1))
a = 15 - (0.5 × 2)
a = 15 - 1
a = 14
Therefore, the first term of the AP is 14.
To find the sum of the first ten terms of the AP, we can use the formula for the sum of an arithmetic progression:
sum of n terms (Sn) = (n/2) × (2a + (n-1)d)
Using the formula with the given values:
Sn = (10/2) × (2(14) + (10-1)(0.5))
Sn = 5 × (28 + 9(0.5))
Sn = 5 × (28 + 4.5)
Sn = 5 × 32.5
Sn = 162.5
Therefore, the sum of the first ten terms of the AP is 162.5.
To find the common difference (d), we can use the formula for the nth term of an arithmetic progression (AP):
an = a1 + (n - 1) * d
Where:
an = nth term of the AP
a1 = first term of the AP
d = common difference of the AP
n = position of the term in the AP
We are given that the 3rd term (a3) is 15 and the 15th term (a15) is 21.
Using the formula, we can create two equations:
a3 = a1 + (3 - 1) * d (Equation 1)
a15 = a1 + (15 - 1) * d (Equation 2)
Substituting the given values:
15 = a1 + 2d (Equation 1)
21 = a1 + 14d (Equation 2)
To solve these equations, we will eliminate variable a1 from the equations:
Multiply Equation 1 by 7:
15 * 7 = 7a1 + 14d
105 = 7a1 + 14d (Equation 3)
Subtract Equation 3 from Equation 2:
21 - 105 = a1 + 14d - 7a1 - 14d
-84 = -6a1
Solving for a1:
-84 / -6 = a1
a1 = 14
Now that we have the value of a1, we can substitute it back into Equation 1 or Equation 2 to find the common difference (d). Let's use Equation 1:
15 = 14 + 2d
Subtract 14 from both sides:
15 - 14 = 14 + 2d - 14
1 = 2d
Divide by 2:
1 / 2 = 2d / 2
1/2 = d
Therefore, the common difference (d) is 1/2.
To find the sum of the first ten terms, we can use the formula for the sum of an arithmetic progression:
Sn = (n/2)(2a1 + (n-1)d)
Substituting the values:
n = 10 (first ten terms)
a1 = 14 (first term)
d = 1/2 (common difference)
Sn = (10/2)(2 * 14 + (10-1) * 1/2)
= 5(28 + 9/2)
= 5(28 + 4.5)
= 5(32.5)
= 162.5
Therefore, the sum of the first ten terms is 162.5.