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.
CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 13(ii)
Given series is an A.P. with first term(a) = 3,
Common difference(d) = 11 − 3 = 8 and nth term(an) = 803.
We know nth term of an A.P. is given by, an = a + (n − 1)d.
=> 803 = 3 + (n − 1)8
=> 803 = 3 + 8n − 8
=> n = 808/8
=> n = 101
Also we know sum of n terms of an A.P. is given by Sn = n[a + an]/2.
S101 = 101[3 + 803]/2
= 101[403]
= 40703
Hence, the sum of terms of the given series is 40703.