if m times the mth term of an AP is equal to the nth times of the nth term and m is not equal to n
show that its (m+n)th term is zero
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dyk:
here given that m is not equal to n
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Hey
Here is your answer
Given m × tm = n × tn
⇒ m[a + (m – 1)d] = n[a + (n – 1)d]
⇒ ma + m(m – 1)d] = na + n(n – 1)d
⇒ ma + m(m – 1)d] – na – n(n – 1)d = 0
⇒ a(m – n) + [m2 – m – n2 + n]d = 0
⇒ a(m – n) + [m2 – n2 – m + n]d = 0
⇒ a(m – n) + [(m – n)(m + n) – (m – n)]d = 0
⇒ (m – n)[a+ {(m + n) – 1}]d = 0
⇒ [a+ {(m + n) – 1}]d = 0
Hence t(m + n) = 0
Hope it helps you!
Here is your answer
Given m × tm = n × tn
⇒ m[a + (m – 1)d] = n[a + (n – 1)d]
⇒ ma + m(m – 1)d] = na + n(n – 1)d
⇒ ma + m(m – 1)d] – na – n(n – 1)d = 0
⇒ a(m – n) + [m2 – m – n2 + n]d = 0
⇒ a(m – n) + [m2 – n2 – m + n]d = 0
⇒ a(m – n) + [(m – n)(m + n) – (m – n)]d = 0
⇒ (m – n)[a+ {(m + n) – 1}]d = 0
⇒ [a+ {(m + n) – 1}]d = 0
Hence t(m + n) = 0
Hope it helps you!
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