A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
If the sum of a certain number of terms of the A.P 25, 22, 19, .... is 116 . Find the last term
Solution:
It is given that 25, 22,19, ....., are in A.P.
Let the sum of n terms of an A.P be 116 .
Here, first term a = 25
Common difference d = 22 - 25 = - 3
We know that, Sn = n/2 [2a + (n - 1) d]
Therefore,
⇒ 116 = n/2 [2(25) + (n - 1) (- 3)]
⇒ 116 = n/2 [50 - 3n + 3]
⇒ 232 = n (53 - 3n) = 53n - 3n2
⇒ 3n2 - 53n + 232 = 0
⇒ 3n2 - 24n - 29n + 232 = 0
⇒ 3n (n - 8) - 29 (n - 8) = 0
⇒ (n - 8)(3n - 29) = 0
⇒ n = 8, 29/3
However, n cannot be equal to 29/3
Hence, n = 8
Therefore, last term l = an = a + (n - 1) d
a8 = 25 + (8 - 1)(- 3)
= 25 + (7)(- 3)
= 25 - 21
= 4
Thus, the last term of the A.P is 4
NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 6
If the sum of a certain number of terms of the A.P 25, 22, 19, .... is 116 . Find the last term
Summary:
The last term of the A.P 25, 22, 19... where the sum is 116 is 4
Math worksheets and
visual curriculum
visual curriculum