14 term of an ap is twice its eigth term . 6 term is -8 find sum of its first 20 terms
Answers
_____Here's your Answer _________
Let "a" be the first term and "d" be the common Differences.
A.T.Q.
a14 = 2(a8)
=> a + 17d = 2 ( a + 7d)
=> a + 17d = 2a + 14d
=> 2a - a = 17d - 14d
=> a = 3d ------------------(1)
Now,
a6 = -8
=> a + 5d = -8
putting eq 1 here, we get
=> 3d + 5d = -8
=> 8d = -8
=> d = -1.------------(2)
Putting eq 2 in equation 1. we get,
a = 3d
=> a = -3.
Now,
s20 = 20/2 [ 2a + ( n-1)d]
=> 10 [ -6 + 19 × (-1)]
=> 10 [ -6 -19]
=> 10 × (- 25)
=> -250
✔✔✔
➡Let the first term of the A.P be a.
➡Let the common difference be d.
We know that the n th term of an A.P is a + ( n - 1 ) d
14 th term = a + ( 14 - 1 ) d
= a + 13 d
8 th term = a + ( 8 - 1 ) d
= a + 7 d
Given : 14 th term = 2 × 8 th term
➡ a + 13 d = 2 ( a + 7 d )
➡ a + 13 d = 2 a + 14 d
➡ a - 2 a + 13 d - 14 d = 0
➡ - a - d = 0
➡ a + d = 0 ...........................( 1 )
Also it is given that 6 th term is - 8
➡ a + ( 6 - 1 ) d = - 8
➡ a + 5 d = - 8 ..............................( 2 )
Subtracting ( 2 ) from ( 1 ) we get :
➡ d - 5 d = 0 - ( - 8 )
➡ - 4 d = 8
➡ d = 8 / - 4
➡ d = - 2
Also a + d = 0
a + ( - 2 ) = 0
➡ a - 2 = 0
➡ a = 2
Hence we can find the sum of 20 terms easily .
sum of n terms = n / 2 × [ 2 a + ( n - 1 ) d ]
Put n = 20 :
➡ 20 / 2 × [ 2 a + ( 20 - 1 ) d ]
➡ 10 × [ 2 × 2 + 19 × ( - 2 ) ]
➡ 10 × [ 4 - 38 ]
➡ 10 × [ - 34 ]
➡ - 340
The sum of the first 20 terms is - 340