Arithmetic
Progressions
Q.3] The sum of first m terms of an AP is same as the sum of its n terms, show that the sum of its (m + n) term is zero. (m - n ≠ 0)
Sol:
Sm
=
Sn
We need to show that:
S(m + n) = 0
=
m
[2a
+ md
– d ]
=
n
[2a
+ nd
– d ]
2am
+ m2d
– md
=
2an
+ n2d
– nd
We need ‘0’ in RHS
2am
– 2an
+ m2d
– n2d
– md
+ nd
= 0
2a
Take 2a common
(m – n)
Take d common
+ d
(m2 – n2)
Take -d common
– d
(m – n)
= 0
Bring all terms to LHS
How to get (m – n) in this term?
By using
a2 – b2 = (a – b)(a + b)
2a(m – n)
+ d
(m + n)(m – n)
– d(m – n)
= 0
Take (m – n) common
(m – n)
[2a
+ d(m + n)
– d]
= 0
2a + d(m + n) – d
= 0
S(m+n)
2a + dm + dn – d
= 0
….(i)
[2a + dm + dn – d]
[0]
…(From i)
S(m+n) = 0
The sum of (m + n)th term is 0
S(m + n)
For Sm,
replace n by m
Additional example