Question

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer 1

Solution:
We have given \(a_{3}=16\)
\(a+(3-1) d=16\)
\(a+2 d=16(1)\)
\(a_{7}-a_{5}=12\)
\([o-(7-1) d]-[a+(5-1) d]=12\)
\((a+6 d)-(a+4 d)=12\)
\(2 d=12\)
\(d=6\)
From equation (1), we obtain
\(a+2(6)=16\)
\(a+12=16\)
\(a=4\)
Therefore, A.P. will be
\(4,10,16,22, \ldots\)

Answer 2

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Answer 3

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