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?

Answer 1

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.