CBSE Class 10 Answered
Expert Answer
Let a and d be the first term and common diff for the A.P.
acc to qn
[a + [(m + 1) - 1]d = 2[a + (n + 1) - 1]d
so
a + md = 2a + 2nd
so
a = (m - 2n)d...(i)
now,
3m+1 th term
=a + [(3m + 1) - 1]d
=a + 3md...
=4md-2nd (ii) using(I)
twice the (m + n + 1)th term
=2{[a + (m + n + 1) - 1]d}
=2[a + (m + n)d]
=2[(m - 2n)d + (m + n)d]
=4md - 2nd..(iii)
from (ii) and (iii) the result is proved.
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