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प्रश्न
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
उत्तर
tn = a + (n – 1)d
tm+1 = a + (m + 1 – 1)d
= a + md
tn+1 = a + (n + 1 – 1)d
= a + nd
2(tn+1) = 2(a + nd)
tm+1 = 2tn+1 ...(1)
⇒ a + md = 2(a + nd)
2a + 2nd – a – md = 0
a + (2n – m)d = 0
t(3m+1) = a + (3m + 1 – 1)d
= a + 3md
t(m+n+1) = a + (m + n + 1 – 1)d
= a + (m + n)d
`2("t"_("m"+ "n" + 1))` = 2(a + (m + n)d)
= 2a + 2md + 2nd
`"t"_((3"m" + 1)) = 2"t"_(("m" + "n" + 1))` ...(2)
a + 3md = 2a + 2md + 2nd
2a + 2md + 2nd – a – 3md = 0
a – md + 2nd = 0
a + (2n – m)d = 0
∴ It is proved that `"t"_((3"m" + 1)) = 2"t"_(("m" + "n" + 1))`
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