Advertisements
Advertisements
Question
How many terms of the A.P. 27, 24, 21, ..... must be taken so that their sum is 105? Which term of the A.P. is zero?
Sum
Solution
Given that,
AP 27, 24, 21, .....
For the given AP,
First term, a = 27
Common difference, d = 24 − 27 = −3
Sum, S = 105 ...[Given]
Let n number of terms be taken for sum to be 105
So, we have,
105 = `n/2[2 xx 27 + (n − 1)(−3)]`
210 = n[54 − 3n + 3]
210 = n(57 − 3n)
210 = 57n − 3n2
3n2 − 57n + 210 = 0
n2 − 19n + 70 = 0
⇒ n2 − 14n − 5n + 70 = 0
⇒ n(n − 14) − 5(n − 14) = 0
⇒ (n − 14)(n − 5) = 0
⇒ n = 5, 14
So, 5 and 14 terms of the given AP must be taken to get sum as 105.
Now,
Let pth term of the AP be zero
⇒ ap = 0
⇒ a + (p − 1)d = 0
⇒ 27 + (p − 1)(−3) = 0
⇒ 27 − 3p + 3 = 0
⇒ 30 = 3p
⇒ p = 10
Hence, 10th term of the given AP is zero.
shaalaa.com
Is there an error in this question or solution?