Question

is an A.P. p, q are distinct natural numbers such that tp = q and tq = p. Find the value of tp+q

Answer 1

Let us take that the first term of the AP is a and the common difference is d

Given that,

$t_{p}$ = q

⇨ a + (p - 1)d = q ...(i)

and

$t_{q}$ = p

⇨ a + (q - 1)d = p ...(ii)

Now, (i) - (ii) ⇨

(p - 1)d - (q - 1)d = q - p

⇨ (p - 1 - q + 1)d = q - p

⇨ (p - q)d = - (p - q)

⇨ d = - 1

Now, putting d = - 1 in (i), we get

a + (q - 1) (- 1) = p

⇨ a = p + q - 1

Hence, the m-th term be

$t_{m}$= a + (m - 1)d

= p + q - 1 + (m - 1) (- 1)

= p + q - 1 - m + 1

= p + q - m

Hence, proved.

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