Q. 36

# If the mth term of an A.P. is and the nth term is , then prove that the sum to mn terms is , where in m n.

Given: Now, am = a + (m – 1)d an + n(m – 1)d = 1

an + mnd – nd = 1 …(i)  am + mnd – md = 1 …(ii)

From eq. (i) and (ii), we get

an + mnd – nd = am + mnd – md

a(n – m) –d (n – m) = 0

a = d

Now, putting the value of a in eq. (i), we get

dn + mnd – nd = 1

mnd = 1 Hence, Sum of mn terms of AP is     Hence Proved

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