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Question
The 4th term of an A.P. is 22 and the 15th term is 66. Find the first terns and the common
difference. Hence find the sum of the series to 8 terms.
Solution
Let a be the first term and d be the common difference of given A.P.
Now,
4th term = 22
⇒ a + 3d = 22 ...(i)
15th term = 66
⇒ a + 14d = 66
Subtracting (i) from (ii), we have
11d = 44
⇒ d = 4
Substituting the value of d in (1) we get
a = 22 - 3 x 4 = 22 - 12 =10
⇒ First term = 10
Now
Sum of 8 terms = `8/2[2xx10+7xx4]` = 4[20+28] = 4x48=192
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