if an A. P. consists of n terms with first term a and n th term I show that the sum of the n th term from the beginning and the m th term from the end is (a + I)
Answers
Step-by-step explanation:
Given :-
In an A. P. consists of n terms with first term a and n th term I .
To find :-
Show that the sum of the n th term from the beginning and the m th term from the end is (a + I)
Solution :-
Given that :
First term in an AP = a
Number of terms in the AP = n
Let the common difference of the AP be d
We know that
nth term of an AP = a+(n-1)d
nth term of the AP from the beginning
an = l
l = a+(n-1)d --------(1)
The AP from the beginning
= a , a+d , a+2d, ..., a+(n-1)d
The AP from the end =
[a+(n-1)d], [a+(n-2)d], ..., a+2d, a+d ,a
=> l, [a+(n-2)d], ..., a+2d, a+d ,a
mth term= l -(m-1)(d )------(2)
Given that there are n terms then
mth term = nth term = l-(n-1)d
=> l-nd+d
We have
nth term of the AP from the beginning = l
nth term of the AP from the end = a
On adding (1),(2)
(1)+(2) =>
Their sum = a+(n-1)d+l-nd+d
=> a+nd-d+l-nd+d
=> a+l
Hence, Proved.
Answer:-
The sum of the n th term from the beginning and the m th term from the end is (a + I)
Used formulae:-
The general form of an AP = a,a+d,...,a+(n-1)d
a = First term
d = Common difference
n = Number of terms
an = a+(n-1)d
an = nth term