Question

If m times the mth term of an AP is equal to n times its nth term then show
that (m+n)th term of an AP is 0
only answer the question those who know

Answer 1

Step-by-step explanation:

Let the first term be 'a' and common difference be 'd'.

m times the mth term of an AP is equal to n times its nth term

=> m x [a + (m - 1)d] = n x [a + (n - 1)d]

=> m(a + md - d) = n(a + nd - d)

=> am + m²d - md = an + n²d - nd

=> am + m²d - md - an - n²d + nd = 0

=> a(m - n) + d(m² - n²) - d(m - n) = 0

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

=> (m - n){a + d(m + n) - d} = 0

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

Note that (m+n)th term is a + (m+n-1)d which has been proved 0.

Proved.

Answer 2

Answer:

Step-by-step explanation

mth term = a+ (m-1)d

nth term = a+(n-1)d

$m [a+ (m-1)d ] = n [ a + (n-1)d ]\\am + m^2d-md = an + n^2d - nd\\am - an + m^2d - n^2d - md + nd = 0\\a(m-n) + d(m^2-n^2) - d(m-n) = 0\\a(m-n) + d(m+n)(m-n) - d(m-n) = 0\\m-n [ a + d(m+n) - d ] = 0\\a + d(m+n) - d = 0\\a +d(m+n-1)= 0\\(m+n) th term = a +(m+n-1)d = 0$