The 1st and the last term of an AP are –4 and 146. The sum of th...
Sn = n/2 ( a +b ) where a is first term and b is last term .
Now, 7171 = n/2 ( -4 +146)
14342 = n × 142
n = 14342 / 142
n = 101
The 1st and the last term of an AP are –4 and 146. The sum of th...
Given information:
- First term of AP = 4
- Last term of AP = 146
- Sum of terms = 7171
To find:
- Number of terms in the AP
Solution:
Let the number of terms in the AP be n.
Using the formula for the sum of an AP, we can write:
Sum of AP = (n/2) x [2a + (n-1)d]
where a is the first term, d is the common difference, and n is the number of terms.
Substituting the given values, we get:
7171 = (n/2) x [2x4 + (n-1)d]
7171 = n/2 x [8 + (n-1)d]
Dividing both sides by n/2, we get:
14342/n = 8 + (n-1)d
Rearranging, we get:
d = [14342/n - 8]/(n-1)
We also know that the first term a = 4 and the last term l = 146.
Using the formula for the nth term of an AP, we can write:
l = a + (n-1)d
Substituting the values, we get:
146 = 4 + (n-1)d
142 = (n-1)d
Dividing both sides by n-1, we get:
d = 142/(n-1)
Equating the two expressions for d, we get:
[14342/n - 8]/(n-1) = 142/(n-1)
Multiplying both sides by (n-1), we get:
14342/n - 8 = 142
Simplifying, we get:
n = 101
Therefore, the number of terms in the AP is 101.
Answer: (a) 101