An AP consist of 21 terms The sum of three term in the middle is 63 and sum of last three terms is 90 Find AP
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The three middle terms are 10th, 11th and 12 th
According to question
T10 + T11 +T12 = 63
=> a+9d +a +10d +a +11d = 63
=> 3a +30d = 63
=> a+10d = 21 -------(1)
Now,
T19 + T20 +T21 = 90
=> a+18d +a +19d +a + 20d = 90
=> 3a +57d = 90
=> a +19d = 30 --------(2)
On subtracting equation 1 from 2, we get
9d = 9
d = 1
a = 11
Required AP is 11, 12, 13,.........upto 21 terms
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