The no of terms in an AP is 40 and the last is -54,given that the sum of the 15 terms added to the sum of the first 30 terms is zero.calculate(1)the first term and common difference.(2)sum of progression
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Step 1: Find the common difference (d).
To find the common difference (d), we can use the formula for the nth term of an arithmetic progression (AP):
nth term = a + (n-1)d
We know that the 40th term is -54, so we can substitute the values into the formula:
-54 = a + (40-1)d
-54 = a + 39d
Step 2: Calculate the sum of the 15 terms.
To find the sum of the first 15 terms, we use the formula:
Sum of n terms = (n/2)(2a + (n-1)d)
Substituting the values given:
Sum of 15 terms = (15/2)(2a + (15-1)d)
Step 3: Calculate the sum of the first 30 terms.
Using the same formula for the sum of n terms, we can find the sum of the first 30 terms:
Sum of 30 terms = (30/2)(2a + (30-1)d)
Step 4: Equate the sum of the 15 terms and the sum of the first 30 terms to zero.
Since the sum of the 15 terms added to the sum of the first 30 terms is zero, we can set up the equation:
Sum of 15 terms + Sum of 30 terms = 0
Simplifying the equation, we get:
(15/2)(2a + (15-1)d) + (30/2)(2a + (30-1)d) = 0
Now we have two equations, one for the common difference (d) and one for the first term (a). We can solve these equations simultaneously to find the values of a and d.
To find the first term and common difference of the arithmetic progression (AP), we can use the formulas:
(1) The nth term of an AP:
an = a1 + (n - 1)d
(2) Sum of the first n terms of an AP:
Sn = n/2 * (a1 + an)
Given information:
- Number of terms (n) = 40
- Last term (an) = -54
- The sum of the first 15 terms (S15) + the sum of the first 30 terms (S30) = 0
Let's solve the problem step by step:
(1) From the given information, we know that the last term (an) is -54. Using the formula for the nth term, we can substitute the values and solve for the first term (a1) and the common difference (d):
an = a1 + (n - 1)d
-54 = a1 + (40 - 1)d
-54 = a1 + 39d
(2) The sum of the first 15 terms (S15) + the sum of the first 30 terms (S30) = 0. Using the formula for the sum of the first n terms, we can substitute the values and solve for the unknowns:
S15 + S30 = 0
15/2 * (a1 + a15) + 30/2 * (a1 + a30) = 0
Now, let's substitute the formula for the nth term into the sum equation and solve for a1 and d simultaneously:
15/2 * (a1 + a1 + (15 - 1)d) + 30/2 * (a1 + a1 + (30 - 1)d) = 0
15/2 * (2a1 + 14d) + 30/2 * (2a1 + 29d) = 0
15(2a1 + 14d) + 30(2a1 + 29d) = 0
30a1 + 210d + 60a1 + 870d = 0
90a1 + 1080d = 0
a1 + 12d = 0 (Dividing by 90)
Now we have two equations:
-54 = a1 + 39d
a1 + 12d = 0
Solve these two equations to find the values of a1 and d.
a+39d = -54
S15 = 15/2 (2a+14d)
S30 = 30/2 (2a+29d)
So, 15/2 (2a+14d) + 30/2 (2a+29d) = 0
or, a+12d = 0
Now you have two equations, and you can find a and d, and then the sum S40.