Question

If 14th term of ap is twice its 8th term and its 6th tetm is -8 then find the sum of its 1st 20tetms

Answer 1

Given that
$a + 13d = 2(a + 7d) \\ a + 13d = 2a + 14d \\ a + d = 0$
And
$a + 5d = - 8 \\ a = - d \\ 4d = - 8 \\ d = - 2$
$a = 2$
$s = \frac{n}{2} (2a + (n - 1)d) \\ s = 10(4 - 38) \\ s = - 340$

Answer 2

Heya !!

Given, 14th term = 2 ( 8th term )

=> a+13d = 2 (a+7d)

=> a+13d = 2a+14d

=> 2a–a = –14d+13d

=> a = –d

And 6th term = –8

=> a+5d = –8

Substituting a = –d in this equation

=> a+ (–5a) = –8

=> –4a = –8

=> a = 8 / 4

=> a = 2

So, d = –2

Sum of 20 terms = n/2 [ 2a + ( n–1) d ]

=> 20/2 [ 2(2) + (20–1) –2 ]

=> 10 [ 4 + 19 (–2) ]

=> 10 [ 4 – 38 ]

=> 10 × –34

=> –340