If (m +1)th term of an AP is twice
the (n +1)th term, then the (3m +1)th
term is twice the
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Answer:-
Given:-
(m + 1)th term of an AP = 2 * (n + 1)th term.
We know that,
nth term of an AP – a(n) = a + (n - 1)d
Hence,
→ a + (m + 1 - 1)d = 2 * [ a + (n + 1 - 1)d ]
→ a + md = 2 (a + nd)
→ a + md = 2a + 2nd
→ md - 2nd = 2a - a
→ a = md - 2nd -- equation (1).
Now,
a(3m + 1) = 2 * a(n)
[ let the term be n]
→ a + (3m + 1 - 1)d = 2 * [ a + (n - 1)d ]
→ a + 3md = 2 * [ a + (n - 1)d ]
Substitute a value from equation (1).
→ md - 2nd + 3md = 2 * [ a + (n - 1)d ]
→ 4md - 2nd = 2 * [ a + (n - 1)d ]
→ 2(2md - nd) = 2 * (a + nd - d)
→ 2md - nd - a + d = nd
→ 2md - nd - (md - 2nd) + d = nd
[ From equation (1) ]
→ 2md - nd - md + 2nd + d = nd
→ md + nd + d = nd
→ (m + n + 1)d = n * d
→ (m + n + 1) = n
Hence, the (3m + 1)th term is twice the (m + n + 1)th term of given AP.
amitkumar44481:
Perfect :-)
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