If (m + 1)h term of an A. P. is twice the (n + 1)th term. Prove that the (3m + 1)th term is twice
the (m +n+ 1)th term.
Answers
Answer:
2am+n+1
Step-by-step explanation:
Given (m+1)
th
term of an AP is twice of (n+1)
th
term, Let a be first term & d be common difference a
m+1
=2a
n+1
a+(m+1−1)d=2[a+(n+1−1)d]
a+md=2a+2nd
a+md=2a+2nd
d(m−2n)=a
d=
m−2n
a
Now (3m+1)
th
term of AP is
a
3m+1
=a+(3m+1−1)d=a+3md
putting values of d we get
a
3m+1
=a+3m(
m−2n
a
)=
m−2n
am−2am+3am
a
3m+1
=
m−2n
4am−2an
……..(1)
Now (m+n+1)
th
term of AP is
a
m+n+1
=a+(m+n+1−1)d
=a+(m+n)d=a+
m−2n
(m+n)a
a
m+n+1
=
m−2n
am−2an+am+an
a
m+n+1
=
m−2n
2am−an
………..(2)
on comparing equation (1) & (2) we get
[a
3m+1
=2
am+n+1
]
let a be the first term and d the common difference of the AP
t(m+1) = a + (m+1-1)d
= a + md
t(n+1) = a + nd
given
t(m+1) = 2 t(n+1)
a + md = 2(a + nd)
a + md = 2a + 2nd
2a - a = md - 2nd
a = (m - 2n) d...........(1)
now
t(3m+1) = a + (3m + 1 - 1)d
= a + 3md
= a + md + 2md
= a + md - 2nd + 2nd + 2md
= a + (m - 2n)d + 2d(m + n)
= a + a + 2d(m + n). ( from (1))
= 2a + 2d(m+n)
= 2{a + (m+n)d}
= 2 { a + (m + n + 1 - 1)d}
= 2 t(m+n+1)
proved