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Question
If 5th and 6th terms of an A.P. are respectively 6 and 5, find the 11th term of the A.P.
Solution
The general term of an A.P. is given by
tn = a + (n – 1)d
Now, t5 = 6
`=>` a + (5 – 1)d = 6
`=>` a + 4d = 6 ...(i)
And t6 = 5
`=>` a + (6 – 1)d = 5
`=>` a + 5d = 5 ...(ii)
Subtracting (ii) from (i), we get
– d = 1
`=>` d = –1
Substituting d = –1 in (i), we get
a + 4(–1) = 6
`=>` a – 4 = 6
`=>` a = 10
`=>` tn = 10 + (n – 1)(–1)
`=>` t11 = 10 + (11 – 1)(–1)
= 10 – 10
= 0
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