QUESTION 2
Easy
Determine the AP whose third term is 16 and whose \(7^{th}\) term exceeds the \(5^{th}\) term by 12.
SOLUTION
1
Form equations
Third term, \(T_3 = 16\)
⇒ \(a + 2d = 16\) ---(I)
Fifth term: \(a + 4d\)
Seventh term: \(a + 6d\)
Given:
\((a + 6d) − (a + 4d) = 12\)
⇒ \(2d = 12\)
⇒ \(d = 6\)
⇒ \(a + 2d = 16\) ---(I)
Fifth term: \(a + 4d\)
Seventh term: \(a + 6d\)
Given:
\((a + 6d) − (a + 4d) = 12\)
⇒ \(2d = 12\)
⇒ \(d = 6\)
2
Find the first term
Substituting \(d = 6\) into (I), we get
\(a + 2(6) = 16\)
⇒ \(a + 12 = 16\)
⇒ \(a = 4\)
\(a + 2(6) = 16\)
⇒ \(a + 12 = 16\)
⇒ \(a = 4\)
3
Write the AP
The AP is: 4, 4+1(6), 4+2(6), 4+3(6),... = 4, 10, 16, 22, 28,...
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Final Answer : The AP is: 4, 10, 16, 22, 28,...
Concept Note
The difference between two terms of an AP depends ont eh number of positions between them.
\(T_7 − T_5 = (a + 6d) − (a + 4d) = 2d\)
This often helps find \(d\) directly.