Question

if the sum of m term of an ap is the same as the sum of its and term show that the sum of its M + n term is zero​

Answer 1

Answer:

Step-by-step explanation:

Sum of m terms=Sum of n terms

=> m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)

Cancelling 2 in the denominator on both sides,

we get

m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

2am + m^2d - md -2an -n^2d +nd =0

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

Taking (m-n) common

2a + ( m + n -1) d = 0 ------------ (1)

S m+n = m+n/2( 2a + (m+n -1)d

we know that 2a + (m+n)d is 0 from eqn. 1

therefore S m+n = 0

Answer 2

Step-by-step explanation:

$\frac{n}{2} (2a + (n - 1)d) = \frac{m}{2} (2a + (m - 1)d) \\ (m - n)a + ( {m}^{2} - {n}^{2} )d - (m - n)d = 0 \\ (m - n)(a + m + n - d) = 0 \\ m - n = 0 :/ m+n=d -a$

Can you proceed .