Question

If the pth term of an A.P. is q and the qth term is p. Prove that its nth term is (p+q-n). ​

Answer 1

Answer:-

Given:

pth term of an AP = q

qth term = p

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

Hence,

⟹ a + (p - 1)d = q

⟹ a + pd - d = q

⟹ a = q - pd + d -- equation (1)

Similarly,

⟹ a + (q - 1)d = p

substitute the value of a from equation (1).

⟹ q - pd + d + qd - d = p

⟹ qd - pd = p - q

⟹ - d(p - q) = p - q

⟹ - d = 1

⟹ d = - 1

Substitute the value of d in equation (1).

⟹ a = q - p( - 1) + ( - 1)

⟹ a = q + p - 1

Now,

aₙ = q + p - 1 + (n - 1)( - 1)

⟹ aₙ = q + p - 1 - n + 1

⟹ aₙ = p + q - n

Hence, Proved.

Answer 2

Given : -

• If the pth term of an A.P. is q and the qth term is p.

To prove:

• nth term of AP is q + p - n.

Solution : -

let,

• First term = a
• Common difference = d

We know Formula of a term of AP,

aⁿ = a+(n - 1 )d

According to the question,

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

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

Subtract eqn (2) from (1)

we get,

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

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

d ( p - q ) = q - p

d = -1

Now,

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

a - p + 1 = q

a = q + p - 1

Therefore,

nth term given by,

aⁿ =(q + p - 1)+(n - 1)( - 1) = q + p - 1 - n + 1 = q + p - n

Hence Proved.