Question

If pth term of an ap is q and qth term is p then show that its nth term is (p+q-n)

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{In an A.P}$

$\mathsf{t_p=q\;\;\&\;\;t_q=p}$

$\underline{\textsf{To prove:}}$

$\mathsf{t_n=p+q-n}$

$\underline{\textsf{Solution:}}$

$\underline{\textsf{Formula used:}}$

$\textsf{The n th term of A.P a, a+d, a+2d,....... is}$

$\mathsf{t_n=a+(n-1)d}$

$\mathsf{t_p=q}$

$\implies\mathsf{a+(p-1)d=q}$....(1)

$\mathsf{t_q=p}$

$\implies\mathsf{a+(q-1)d=p}$....(2)

$\textsf{Solving (1) and (2)}$

$\mathsf{a+(p-1)d=q}$

$\mathsf{a+(q-1)d=p}$

$\textsf{Subtracting}$

$\mathsf{pd-qd=q-p}$

$\mathsf{(p-q)d=-(p-q)}$

$\implies\boxed{\mathsf{d=-1}}$

$\textsf{From (1)}$

$\mathsf{a+(p-1)(-1)=q}$

$\mathsf{a+1-p=q}$

$\implies\boxed{\mathsf{a=p+q-1}}$

$\textsf{Now}$

$\mathsf{t_n=a+(n-1)d}$

$\mathsf{t_n=(p+q-1)+(n-1)(-1)}$

$\mathsf{t_n=p+q-1+1-n}$

$\implies\boxed{\mathsf{t_n=p+q-n}}$

$\textbf{Find more:}$

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Answer 2

nth term = (p+q-n)

Step-by-step explanation:

In an AP,

pth term = q, qth term = p

To prove,

nth term = (p+q-n)

$t_{p}$ = q

⇒ a+(p−1)d = q  ...(1)

$t_{q}$ = p

⇒ a+(q−1)d = p  ...(2)

Solving these equations, we get,

pd - qd = q - p

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

⇒ d=−1

From equation (1),

a = (p+q−1)

Thus,

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

$t_{n}$ = (p + q - 1) + (n - 1) (-1)

$t_{n}$ = P + q - 1 + 1 - n

∵ nth term or $t_{n}$ = (p+q−n)

Learn more: Arithmetic progressions

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