This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.6
In this question we have been given an arithmetic progression. And we have to find its sum of last terms.
CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 4
Given A.P. has first term(a) = 8,
Common difference(d) = 10 – 8 = 2
and nth term(an) = 126.
The nth term of the A.P. is given by, an = a+(n – 1)d
=> 126 = 8 + (n – 1)(2)
=> 126 = 8 + 2n – 2
=> 2n = 120
=> n = 60
We have to find the sum of last ten terms i.e.,
S = a51 + a52 + a53 + ……. + a60
a51 = 8 + (51 – 1)(2) = 8 + 50(2) = 108
So, sum would be S = 10[108 + 126]/2 = 5(234) = 1170
Hence, the sum of last 10 terms of the A.P. is 1170.