If an ap consists of m terms with first term a and nth term l show that
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the sum of mth term from the begining and the mth term from the end is (a+L
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Given: First term of the AP = a
Given : Last term of the AP = 1.
Given : Total number of terms = n.
Common difference from the beginning=d
Common difference from the end = -d.
.. Last term of the AP = nth term of the AP
» an = |
» a + (n - 1) * d = 1
Now,
→nth term of the AP from the beginning an = a + (n - 1) * d
→ nth term from the end = 1 + (n - 1) * (-d)
= 1 - (n - 1) * d
Sum of nth term from the beginning + nth term from the end is given by:
→a + (n - 1) * d + [1 - (n - 1) * d]
→ a + (n - 1) * d+1- (n - 1) * d
→ a + nd - d +1- nd + d
→a + l
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