if (m+1)th term of an A.P. is twice the (n+1)th term, then prove that (3m+1)th term is twice the (m+n+1)th term.
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Answers
Answer:
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
].
solution
Answer:
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
].
Step-by-step explanation:
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