Math, asked by tajju, 1 year ago

how many terms of the A.P 18, 16, be taken so that their sum is zero

Answers

Answered by triptig
5
if you find it helpful please mark it as brainlist
sorry the above answer is wrong. the correct one is below

Sn=n/2(2a+(n-1)d)
0=2*18+(n-1)(-2)
0=36-2n+2
0=38-2n
2n=38
n=19
Attachments:

bishista: Umm.... You are wrong my dear. It is the sum not the term.
triptig: yes thats why i have corrected my mistake
bishista: Yup. Now its alright
Answered by Anonymous
1

Answer:

Consider  \: the \:   \: given \:  A.P.  \: series.</p><p> \\ </p><p>27,24,21,......</p><p></p><p> \\ </p><p>Here, a=27,d=−3</p><p></p><p> \\ </p><p>Since, Sum=0</p><p> \\ </p><p></p><p>Therefore,</p><p></p><p>

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

0=  \frac{n}{2} [2 \times 27 + (n - 1) \times  - 3]

54 - 3n  + 3 = 0

57 - 3n = 0

57 = 3n

n =  \frac{ \cancel{ 57}}{ \cancel 3}  = 19

so, \:  \boxed{n = 19}

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