Advertisements
Advertisements
प्रश्न
How many terms of the A.P. 27, 24, 21, ..., should be taken so that their sum is zero?
Advertisements
उत्तर
A.P. = 27, 24, 21,...
a = 27
d = 24 – 27 = –3
Sn = 0
Let n terms be there in A.P.
Sn = `n/(2)[2a + (n -1)d]`
⇒ 0 = `n/(2)[(2 xx 27) + (n - 1)(-3)]`
⇒ 0 = n[54 – 3n + 3]
⇒ n[57 –3n] = 0
⇒ (57 – 3n) = `(0)/n` = 0
⇒ 3n = 57
∴ n = `(57)/(3)`
= 19
संबंधित प्रश्न
The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of numbers of houses preceding the house numbered x is equal to sum of the numbers of houses following x.
Find the sum of the following APs:
–37, –33, –29, ... to 12 terms.
The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Find the sum of all 3 – digit natural numbers which are divisible by 13.
How many three-digit natural numbers are divisible by 9?
In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
The common difference of an A.P., the sum of whose n terms is Sn, is
Q.14
If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.
Find the sum of first 16 terms of the A.P. whose nth term is given by an = 5n – 3.
