if the sequence<an> is an A.P show that a(m+n) + a(m-n) = 2am?

Answer 1

Given that the sequence a[n] is an arithmetic progression, that means

a[n] = a[1] + (n - 1)d

It follows that

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

Adding these two equations together,

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

Expanding the right hand side,

a[m + n] + a[m - n] = a[1] + md + nd - d + a[1] + md - nd - d

Simplifying the right hand side

a[m + n] + a[m - n] = 2a[1] + 2md - 2d (*)

However,

2a[m] = 2 (a[1] + (m - 1)d)
2a[m] = 2 (a[1] + md + d)

2a[m] = 2a[1] + 2md + 2d (**)

Compare equations (*) to (**); they are equal. Therefore, it follows that

a[m + n] + a[m - n] = 2a[m]

Answer 2

a(m+n)=a+(m+n-1)d --------------------1)
a(m-n)=a+(m-n-1)d --------------------2)
a(m)=a+(m-1)d -----------------------3)
add 1) and 2),
a(m+n)+a(m-n)=a+(m+n-1)d+a+(m-n-1)d
=2a+(m+n-1)d+(m-n-1)d
=2a+(m+n-1+m-n-1)d
=2a+(2m-2)d
=2[a+(m-1)d]
=2a(m) -----------------------------------(frm 3)
hence proved.
best of luck for the exams