Given that the first term of A.P is 7 and 10th term is twice the second term,calcute the sum of 28th term ?
a=7
a+9d = 2(a+d)
Solve for a and d.
Now, since a sum requires two addends, I do not know what "the sum of the 28th term" is supposed to be. But, if you meant the sum of the first 28 terms, then
S28 = 28/2 (2a+27d)
So plug and chug.
To calculate the sum of the 28th term of an arithmetic progression (A.P.), we first need to find the common difference.
Given that the first term of the A.P. is 7 and the 10th term is twice the second term, we can write the following equations:
First term: a₁ = 7
10th term: a₁₀ = 2a₂
We know that the formula for the nth term of an A.P. is given by:
aₙ = a₁ + (n - 1)d
Where aₙ is the nth term, a₁ is the first term, n is the term number, and d is the common difference.
Using this formula, we can write the equation for the 10th term:
a₁₀ = 7 + (10 - 1)d
2a₂ = 7 + 9d
Since the 10th term is twice the second term, we can equate these two equations:
7 + 9d = 2a₂
Next, we can substitute the value of a₁ from the first equation (a₁ = 7) into the equation above:
7 + 9d = 2(7 + d)
Simplifying the equation, we get:
7 + 9d = 14 + 2d
7d = 7
d = 1
Therefore, the common difference (d) is 1.
Once we have the value of the common difference, we can calculate the 28th term of the A.P. using the formula:
aₙ = a₁ + (n - 1)d
Substituting the values into the equation:
a₂₈ = 7 + (28 - 1)1
= 7 + 27
= 34
The 28th term of the A.P. is 34.
Now, to calculate the sum of the 28th term, we can use the formula for the sum of an A.P.:
Sₙ = (n/2)(2a₁ + (n - 1)d)
Substituting the values into the equation:
S₂₈ = (28/2)(2 * 7 + (28 - 1) * 1)
= 14(14 + 27)
= 14 * 41
= 574
Therefore, the sum of the 28th term of the A.P. is 574.