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If there are any terms of parallel m range n and n is the post m , then prove that p terms will be m + n-p and (m + n)
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Anonymous:
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Hey Mate
your answer is--
Given, Tm = n
=> a + (m-1) d = n .....(1)
also, given
Tn = m
=> a + (n-1) d = m ....(2)
subtract equation (1) from (2) , we get
a+(n-1)d - { a+(m-1) d} = m - n
=> a+ dn - d - a - dm + d = m - n
=> d ( n - m ) = m - n
=> d = (m-n)/(n-m)
=> d = -(n-m)/(n-m)
=> d = -1 ...(3)
put this value in equation (1), we get
a + (m-1) -1 = n
=> a + (1-m) = n
=> a = n - (1-m) ...(4)
now, for p term
Tp = a + (p-1) d
= n - (1-m) + (p-1) -1
= n - 1 + m - p +1
= n +m - p
Now,
T(m+n) = a + (m+n-1)d
= n-(1-m)-m-n +1
= n -1 + m -m-n+1
= -1 +1 = 0
【 Hope it helps you 】
your answer is--
Given, Tm = n
=> a + (m-1) d = n .....(1)
also, given
Tn = m
=> a + (n-1) d = m ....(2)
subtract equation (1) from (2) , we get
a+(n-1)d - { a+(m-1) d} = m - n
=> a+ dn - d - a - dm + d = m - n
=> d ( n - m ) = m - n
=> d = (m-n)/(n-m)
=> d = -(n-m)/(n-m)
=> d = -1 ...(3)
put this value in equation (1), we get
a + (m-1) -1 = n
=> a + (1-m) = n
=> a = n - (1-m) ...(4)
now, for p term
Tp = a + (p-1) d
= n - (1-m) + (p-1) -1
= n - 1 + m - p +1
= n +m - p
Now,
T(m+n) = a + (m+n-1)d
= n-(1-m)-m-n +1
= n -1 + m -m-n+1
= -1 +1 = 0
【 Hope it helps you 】
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