Question

In a given AP if pth term is q and qth term is p them show that the nth term is (p + q – n).

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Answer 1

Step-by-step explanation:

Given:

• In a AP series.
• pth term of AP is q.
• qth term of AP is p.

To Show:

• nth term of AP is (p + q – n)

Solution: Let a be the first term and d be the common difference of given AP.

As we know that the an AP series is given by

★ a + (n – 1)d ★

∴ pth term will be

➟ ap = a + (p – 1)d

➟ a + (p – 1)d = q.........i

∴ qth term will be

➟ aq = a + (q – 1)d

➟ a + (q – 1)d = p........ii

From equation (i) & (ii) we get

a + (p – 1)d = q

a + (q – 1)d = p

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(p – q)d = (q – p)

Or

➮ d = (q –p)/(p – q)

➮ d = – (–q + p)/(p – q)

➮ d = – (p – q)/(p – q)

➮ d = –1

Put the value of d in equation (i)

$\implies{\rm }$ a + (p – 1)–1 = q

$\implies{\rm }$ a + (–p + 1) = q

$\implies{\rm }$ a – p + 1 = q

$\implies{\rm }$ a = p + q –1

So we got

• a = first term = p + q – 1
• d = common difference = –1

Now nth term of AP will be

$\implies{\rm }$ aⁿ = a + (n – 1)d

$\implies{\rm }$ aⁿ = p + q – 1 + (n – 1)–1

$\implies{\rm }$ aⁿ = p + q – 1 + (–n + 1)

$\implies{\rm }$ aⁿ = p + q – 1 – n + 1

$\implies{\rm }$ aⁿ = p + q – n

Hence, nth term of AP is (p + q – n).

$\large\bold{\texttt {Proved }}$

Answer 2

♧Answer♧

Let a be the first term and d be the common difference of the given AP. Then,

Tp = a+(p-1)d and Tq = a+(q-1)d

Now, Tp = q and Tq = p

• a+(p-1)d = q ........(1)

and a+(q-1)d = p .........(2)

On subtracting 1 from 2, we get

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

d = -1

Putting d = -1 in (1), we get a = (p+q-1)

Thus, a = (p+q-1) and d = -1

• nth term

= a+(n-1)d

= (p+q-1) + (n-1) × (-1)

= p+q-n

Hence, nth term = (p+q-n)

Proved!!