Question -
Answer -
We know the first termis ‘a’ and the common difference of an A.P is d.
Given:
am+n +am–n = 2am
By using the formula,
an = a+ (n – 1)d
Now, let us take LHS:am+n + am-n
am+n +am-n = a + (m + n – 1)d + a + (m – n – 1)d
= a + md + nd – d + a+ md – nd – d
= 2a + 2md – 2d
= 2(a + md – d)
= 2[a + d(m – 1)] {∵ an = a + (n – 1)d}
am+n +am-n = 2am
Hence Proved.