Question

If (r+1)th term of an A.P is twice (p+1)th the term, then prove that (5r+1)th term is twice (2r+p+1)th term

Answer 1

Step-by-step explanation:

Let the first term be 'a'

the common difference is 'd'

Now(AP)

th

term is

=a+(n−1)d

=a+(p+1−1).d

=a+pd−−−−−−(1)

(p+1)

th

=2(q+1)

th

term is

=a+(q+1−1).d

=a+qd−−−−−−(2)

According to question

a+pd=2(a+qd)

a+pd=2a+2qd

a=d[p−2q]−−−−−(3)

Now to prove

(3p+1)

th

term=2×[p+q+1]

th

term

⇒a+(3p+1−1)d.=2×[a+[p+q+1−1]d]

⇒a+3pd=2[a+(p+q)d]

⇒a=3pd−2(p+q).d

⇒a=3pd−2pd−2qd

⇒a=pd−2qd

⇒a=d[p−2q]−−−−(4)

form (3)put the value of a

⇒d[p−2q]=d[p−2q]

∴ L.H.S=R.H.S