The 6th term of an A.P is twice the third and the first term is 3. Find the common difference and the 10th term and find the sum of the 1st 10 terms
To find the common difference in an arithmetic progression (A.P), we need to use the given information: the 6th term is twice the third term.
Let's denote the common difference as 'd'. Since we know the first term is 3, we can write the A.P as:
3, 3 + d, 3 + 2d, ..., 3 + 5d, 3 + 6d, ...
Given that the 6th term (3 + 5d) is twice the third term (3 + 2d), we can write the equation:
3 + 5d = 2(3 + 2d)
Simplifying this equation:
3 + 5d = 6 + 4d
Rearranging terms:
5d - 4d = 6 - 3
d = 3
Hence, the common difference is 3.
Now let's find the 10th term. We already know the first term (a₁) is 3 and the common difference (d) is 3. The nth term of an A.P is given by the formula:
aₙ = a₁ + (n - 1)d
Substituting the values:
a₁₀ = 3 + (10 - 1) × 3
a₁₀ = 3 + 9 × 3
a₁₀ = 3 + 27
a₁₀ = 30
Therefore, the 10th term of the A.P is 30.
To find the sum of the first 10 terms, we will use the formula for the sum of an A.P:
Sₙ = (n/2) × (2a₁ + (n - 1)d)
Substituting the values:
S₁₀ = (10/2) × (2 × 3 + (10 - 1) × 3)
S₁₀ = 5 × (6 + 9 × 3)
S₁₀ = 5 × (6 + 27)
S₁₀ = 5 × 33
S₁₀ = 165
Therefore, the sum of the first 10 terms is 165.
This is just a direct application of your definitions.
"the first term is 3" ---> a = 3
"The 6th term of an A.P is twice the third" ---> a+5d = 2(a+2d)
3 + 5d = 6 + 4d
d = 3
term(10) = a + 9d = ....
sum(10) = (10/2)(2a + 9d) = ...