In an A.P the 10th term is 31 and 15th term is 46. Find the series?
- The 10th term is 31
31 = a + (10 - 1) d
46 = a + (15 - 1) d
Solving these equations a=4 and d=3
Series 4, 7, 10
To find the correct answer, we can use the formula for the nth term of an arithmetic progression (A.P):
an = a1 + (n-1)d
where
an is the nth term of the A.P
a1 is the first term of the A.P
d is the common difference between the terms of the A.P
n is the number of terms in the A.P
Let's use this formula to solve the problem:
Given, a10 = 31 and a15 = 46
Using the formula for the nth term of an A.P, we have:
a10 = a1 + 9d = 31 ----(1)
a15 = a1 + 14d = 46 ----(2)
Subtracting equation (1) from (2), we get:
5d = 15
d = 3
Now, substituting the value of d in equation (1), we get:
a1 + 9(3) = 31
a1 = 4
Therefore, the first term of the A.P is 4 and the common difference is 3.
Using the formula for the nth term of an A.P, we can find the series:
a1 = 4
d = 3
an = a1 + (n-1)d
Using this formula, we get the following series for n = 1 to 3:
a1 = 4
a2 = a1 + d = 7
a3 = a1 + 2d = 10
Therefore, the series is 4, 7, 10, which corresponds to Option A.
In conclusion, the correct answer is Option A, and the series is 4, 7, 10.
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