Question

the 14 term of an ap is twice its 8 term if its 6 term is minus 8 then find the sum of its first 20 term

Answer 1

HELLO DEAR,

given that:-
(1) (t14)=2(t8)

$a + (14- 1)d = 2(a + (8 - 1)d \\ = > a + 13d = 2a + 14d \\ = > a = - d.......(1)$

(2) t6=-8

$a + (6 - 1)d = - 8 \\ = > a +5d = - 8..........(2) \\ = > - d + 5d = - 8......using.(1) \\ = > 4d = - 8 \\ = >d = \frac{ - 8}{4} \\ = > d = - 2$
put in (1)
we get
a= -(-2) = 2

$s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ = > s_{20} = \frac{20}{2} (2 \times 2 + (20 - 1)( - 2)) \\ = > s_{20} = 10(4 + 19 \times ( - 2)) \\ = > s_{20} = 10(4 - 38) \\ = > s_{20} = 10( - 34) \\ = > s_{20} - 340$
I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

I hope the answer is correct
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