How many terms of the AP $27, 24, 21, \dots$ should be taken so that their sum is $0$

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Solution

Consider the given A.P. series.
$27, 24, 21, . . . . . .$

Here, $a = 27, d = - 3$

Since, $S u m = 0$

Therefore,
$S u m = \frac{n}{2} [2 a + (n - 1) d]$
$0 = \frac{n}{2} [2 \times 27 + (n - 1) \times - 3]$
$54 - 3 n + 3 = 0$
$57 - 3 n = 0$
$3 n = 57$
$n = 19$

Hence, this is the answer.

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Q. How many terms of

A P : 27, 24, 21, . . . . .

should be taken so that their sum is zero? What is the value of that last term.

How many terms of the AP 9, 17, 25, ... must be taken so that their sum is 636?

(i) How many terms of the sequence $18, 16, 14, . . .$ should be taken so that their sum is zero?
(ii) How many terms are there in the A.P. whose first and fifth terms are $- 14$ and $2$ respectively and the sum of the terms is $40$ ?
(iii) How many terms of the A.P. $9, 17, 25, . . .$ must be taken so that their sum is $636$ ?
(iv) How many terms of the A.P. $63, 60, 57, . . .$ must be taken so that their sum is $693$ ?
(v) How many terms of the A.P. $27, 24, 21...$ should be taken so that their sum is zero?

(i) How many terms of the sequence $18, 16, 14, . . .$ should be taken so that their sum is zero?
(ii) How many terms are there in the A.P. whose first and fifth terms are $- 14$ and $2$ respectively and the sum of the terms is 40?
(iii) How many terms of the A.P., $9, 17, 25, . . .$ must be taken so that their sum is $636$ ?
(iv) How many terms of the A.P., $63, 60, 57, . . .$ must be taken so that their sum is $693$ ?
(v) How many terms of the A.P., $27, 24, 21...$ should be taken so that their sum is zero?
(vi) How many terms of the A.P., $45, 39, 33...$ must be taken so that their sum is $180$ ?

How many terms of the AP 27,24, 21,..... are required to get the sum as 0?

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How many terms of the AP 27,24,21,… should be taken so that their sum is 0

Consider the given A.P. series.27,24,21,......Here, a=27,d=−3Since, Sum=0Therefore,Sum=n2[2a+(n−1)d]0=n2[2×27+(n−1)×−3]54−3n+3=057−3n=03n=57n=19Hence, this is the answer.

How many terms of the AP $27, 24, 21, \dots$ should be taken so that their sum is $0$

Consider the given A.P. series.
$27, 24, 21, . . . . . .$

Here, $a = 27, d = - 3$

Since, $S u m = 0$

Therefore,
$S u m = \frac{n}{2} [2 a + (n - 1) d]$
$0 = \frac{n}{2} [2 \times 27 + (n - 1) \times - 3]$
$54 - 3 n + 3 = 0$
$57 - 3 n = 0$
$3 n = 57$
$n = 19$

Hence, this is the answer.