Advertisements
Advertisements
Question
How many terms of the A.P. 27, 24, 21, ..., should be taken so that their sum is zero?
Advertisements
Solution
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
RELATED QUESTIONS
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees, and so on till class XII. There are three sections of each class. How many trees will be planted by the students?
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
If (2p – 1), 7, 3p are in AP, find the value of p.
Fill up the boxes and find out the number of terms in the A.P.
1,3,5,....,149 .
Here a = 1 , d =b`[ ], t_n = 149`
tn = a + (n-1) d
∴ 149 =`[ ] ∴149 = 2n - [ ]`
∴ n =`[ ]`
Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is
Q.6
Q.17
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
