The 14th term of an AP is twice the 8th term.If the 6th term is -8 then find the sum of its first 20 terms
To find the sum of the first 20 terms of an arithmetic progression (AP), we need to find the common difference first.
It is given that the 14th term is twice the 8th term. Let's assume that the common difference is denoted by 'd'.
The 8th term can be expressed as the first term plus 7 times the common difference (since it is 7 terms ahead of the 1st term). Similarly, the 14th term can be expressed as the first term plus 13 times the common difference.
Based on the given information, we can set up the following equation:
first term + 13d = 2(first term + 7d)
Since we know the 6th term is -8, we can substitute it into the equation:
first term + 5d = -8
Now, we have a system of two equations:
first term + 13d = 2(first term + 7d)
first term + 5d = -8
To solve this system, we can substitute the value of the first term from the second equation into the first equation:
(-8 - 5d) + 13d = 2((-8 - 5d) + 7d)
Simplifying the equation:
-8 + 8d = -16
Adding 8 to both sides:
8d = -8
Dividing both sides by 8:
d = -1
Next, substitute the value of d back into the second equation to find the first term:
first term + 5(-1) = -8
first term - 5 = -8
Adding 5 to both sides:
first term = -3
Now that we have found the common difference (d = -1) and the first term (first term = -3), we can find the sum of the first 20 terms using the formula:
Sum = (n/2)(2a + (n-1)d)
In this case, n (the number of terms) is 20, a (the first term) is -3, and d (the common difference) is -1.
Sum = (20/2)(2(-3) + (20-1)(-1))
= 10(-6 - 19)
= 10(-25)
= -250
Therefore, the sum of the first 20 terms of this AP is -250.
n th term = a + (n-1) d
14th = a + 13 d = 2 [ 8th] = 2[ a + 7 d ]
so
a + 13 d = 2 a + 14 d
a = -d
6th = a + 5 d = -8
-d + 5 d = -8
d = -2
so a = +2
sum of first n terms = (n/2)[ 2a + (n-1)d]
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= (20/2) [ 2*2 + 19(-2)] =10 [4 - 38] = 10[-34] = -340