Question

The 14 th term of an ap is twice the 8 th term if the 6 th term is -8 then find the sum of its 20 terms

Answer 1

Answer:

lets take the first term as 'a' and let the common difference be d,then given

$a_{14} = 2a_{8}$   -------eq (1)

also,

$a_{14} = a+13d$      and    $a_{8} = a+7d$

then eq(1) becomes

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

a+13d=2a+14d

a-2a=14d-13d

-a=1d  or a= -d ------------eq(2)

also given $a_{6} = -8$

i.e, a+5d= -8  ------- eq (3)

from eq(2) a= -d

then eq (3) becomes

-d + 5d = -8   or   4d= -8 or   d= -8/4  or  d= -2

now,

a= -(-2)  i.e  a=2     d= -2    

we are asked to find the sum of first 20 terms i.e n=20

we have,   $S_{n} = \frac{n}{2} [2a+(n-1)d]$

so,

$S_{20} = \frac{20}{2} [2+(20-1)(-2)]$

     $= 10[2+19(-2)]\\= 10[2-38]\\= 10[-34]\\= -340$

so, sum of 1st 20 terms is -340

hope this helps....