One of the most important and exam oriented question from Chapter name- Arithmetic Progression
Class 10th
Chapter number- 9
Exercise :- 9.6
This type of question has been asked in previous years exams.
In this question we have been given an arithmetic progression such that sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, . . ., is 116.
Now we have to find the last term.
CBSE DHANPAT RAI publication
CBSE Mathematics Class 10th
Question 9
Given A.P. has first term(a) = 25,
Common difference(d) = 22 – 25 = -3 and sum(Sn) = 116.
We know sum of n terms of an A.P. is given by, Sn = n[2a + (n – 1)d] / 2.
=> 116 = n[2(25) + (n − 1)(−3)]/2
=> n[53 − 3n]/2 = 116
=> 53n – 3n2 = 232
=> 3n2 – 53n + 232 = 0
=> 3n2 – 24n – 29n + 232 = 0
=> 3n(n – 8) – 29 (n – 8) = 0
=> (3n – 29)( n – 8 ) = 0
=> n = 29/3 or n = 8
Ignoring n = 29/3 as number of terms cannot be a fraction, so we get, n = 8.
So, the last term is:
a8 = a + (8 – 1)d = 25 + 7(-3) = 25 – 21 = 4
Hence, the last term of the given A.P. is 4.