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Question
The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.
Solution
Let 'a' be the first term and 'd' be the common difference of the given A.P.
Now, t6 = 16 ...(Given)
`\implies` a + 5d = 16 ...(i)
And t14 = 32 ...(Given)
`\implies` a + 13d = 32 ...(ii)
Subtracting (i) from (ii), we get
8d = 16
`\implies` d = 2
`\implies` a + 5(2) = 16
`\implies` a = 6
Hence, 36th term = t36
= a + 35d
= 6 + 35(2)
= 76
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