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Question
The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.
Sum
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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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