Question

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In an arithmetic sequence with common difference 6, ninth term is zero.

Find the sum of first 17 terms?

Answer 1

Answer:

48

Step-by-step explanation:

d=6

ninth term=0

17th term =?

9term=a+8d

put the value of all term

0=a+8*6

0=a+48

a=-48

17th term =a+16d

put the value of all term

17th term= -48+16*6

-48+96

17th term=48

Answer 2

Answer:

0

Step-by-step explanation:

$a_{9} =a+(n-1)d$

 0= a+ (9-1)6

0= a+8×6

0=a+48

a=8

$s_{17} = \frac{n}{2} [2a+(n-1)d]$

    $= \frac{17}{2} [2(-48)+(17-1)6$

    $= \frac{17}{2} [-96+96]$

    $=\frac{17}{2}$*0

    =0

     

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