If m times the n th term of an A.P. is equal to n times its nth term. Then show that m+nth term of A.P. is 0.
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SOLUTION:-
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
SOLUTION:-
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
Answered by
2
✴hy ✴
✴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
✴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
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