The 10 th term of an arithmetic sequence is equal to the sum of 40 and 15th term. If the 15th term is 127, what is the common difference? What is the Ist term ? What is the sum of ist 30 terms?
Let's assume that the first term in the arithmetic sequence is "a" and the common difference is "d".
We know that the 10th term is equal to the sum of 40 and the 15th term:
10th term = 40 + 15th term
Substituting in the value of the 15th term:
10th term = 40 + 127
Simplifying:
10th term = 167
Using the formula for the nth term of an arithmetic sequence, we can find the value of the 10th term:
a_n = a + (n-1)d
10th term = a + (10-1)d
167 = a + 9d
We also know that the 15th term is 127:
a + (15-1)d = 127
a + 14d = 127
Now we have a system of equations:
167 = a + 9d
127 = a + 14d
Subtracting the second equation from the first, we can eliminate the "a" term:
167 - 127 = (a + 9d) - (a + 14d)
40 = -5d
d = -8
So the common difference is -8.
Using the first equation, we can find the value of the first term:
167 = a + 9(-8)
167 = a - 72
a = 239
So the first term is 239.
To find the sum of the first 30 terms, we can use the formula for the sum of an arithmetic sequence:
S_n = (n/2)(a + l)
Substituting in the values:
S_30 = (30/2)(239 + (239 + (30-1)(-8)))
S_30 = 15(239 + 239 + 29(-8))
S_30 = 15(478 - 232)
S_30 = 15(246)
S_30 = 3690
So the sum of the first 30 terms is 3690.
To find the common difference of an arithmetic sequence, we can use the formula:
nth term = a + (n - 1)d
We are given that the 10th term is equal to the sum of 40 and the 15th term. So we can write the equation:
10th term = 40 + 15th term
Substitute the values we know:
10th term = 40 + 127
10th term = 167
Now, use the formula to find the common difference:
167 = a + (10 - 1)d
167 = a + 9d ----- (Equation 1)
We are also given that the 15th term is 127. So we can write the equation:
15th term = a + (15 - 1)d
127 = a + 14d ----- (Equation 2)
Now, we have two equations with two variables. We can solve them simultaneously to find the common difference and the first term.
Let's solve Equation 1 and Equation 2:
Equation 2 - Equation 1:
127 - 167 = (a + 14d) - (a + 9d)
-40 = 14d - 9d
-40 = 5d
Divide both sides by 5:
-40/5 = d
-8 = d
So, the common difference is -8.
To find the first term (a), we can substitute the common difference into one of the equations:
127 = a + 14d
127 = a + 14(-8)
127 = a - 112
a = 127 + 112
a = 239
So, the first term is 239.
To find the sum of the first 30 terms of the arithmetic sequence, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(2a + (n - 1)d)
Substitute the values we know:
n = 30 (the first 30 terms)
a = 239 (the first term)
d = -8 (the common difference)
Sum = (30/2)(2(239) + (30 - 1)(-8))
Sum = 15(478 + 29(-8))
Sum = 15(478 - 232)
Sum = 15(246)
Sum = 3690
So, the sum of the first 30 terms is 3690.