Advertisements
Advertisements
Question
How many terms of the A.P. 27, 24, 21, .... should be taken so that their sum is zero?
Solution
The given AP is 27, 24, 21, ..
First term of the AP = 27
Common difference = 24 − 27 = −3
Let the sum of the first x terms of the AP be 0.
Sum of first x terms = `x/2`[2×27+(x−1)(−3)]=0
⇒`x/2`[54+(−3x+3)]=0
⇒x(54−3x+3)=0
⇒x(57−3x)=0
Now, either x = 0 or 57 − 3x = 0.
Since the number of terms cannot be 0, x≠0.
∴ 57 − 3x = 0
⇒ 57 = 3x
⇒ x = 19
Thus, the sum of the first 19 terms of the AP is 0.
APPEARS IN
RELATED QUESTIONS
Find the sum of all numbers from 50 to 350 which are divisible by 6. Hence find the 15th term of that A.P.
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Which term of the AP 3,8, 13,18,…. Will be 55 more than its 20th term?
Find the three numbers in AP whose sum is 15 and product is 80.
How many three-digit natural numbers are divisible by 9?
The nth term of an AP is given by (−4n + 15). Find the sum of first 20 terms of this AP?
Find the sum of n terms of the series \[\left( 4 - \frac{1}{n} \right) + \left( 4 - \frac{2}{n} \right) + \left( 4 - \frac{3}{n} \right) + . . . . . . . . . .\]
Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d. find A.P.
Find the sum of three-digit natural numbers, which are divisible by 4
Find the sum of the integers between 100 and 200 that are
- divisible by 9
- not divisible by 9
[Hint (ii) : These numbers will be : Total numbers – Total numbers divisible by 9]
