Q. 31

# Find a, b and c i

We have,

Second term, a2 = 7

a + d = 7 [1]

Fourth term, a4 = 23

a + 3d = 23 [2]

[2] – [1] gives,

3d – d = 23 – 7

2d = 16

d = 8

a + 8 = 7

a = -1

b = 7 + d = 7 + 8 = 15

c = 23 + d = 23 + 8 = 31

OR

We know that nth term of A.P. , tn = a + (n – 1) d.

First, tn = a + (n – 1) d

Then, tm = a + (m – 1) d

Given, mtm = ntn

m [a + (m – 1) d] = n [a + (n – 1) d]

ma + m2d – md = na + n2d – nd

ma na + m2d – n2d – md + nd = 0

a (m n) + d (m2 – n2) – d (m – n) = 0

We know that a2 – b2 = (a – b) (a + b)

(m n) [a + d (m + n) d] = 0

[a + d (m + n) d] = 0

a + (m + n 1) d = 0

(m + n)th term, tm + n = 0

Hence proved.

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