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Question
Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
Solution
Let the first term be a and the common difference be d.
It is given that a4 = 9 and a6 + a13 = 40.
a4 = 9
`rArr a+(4-1)d=9` `a_n=a+(n-1)d`
`rArr a+3d=9`
`a_6+a_13=9`
`rArr {a+(6-1)d}+{a+(13-1)d}=40`
`rArr {a+5d}+{a+12d}=40`
`rArr 2a+17d=40`
From (1):
a = 9 − 3d
Substituting the value of a in (2):
2 (9 − 3d) + 17d = 40
⇒ 18 + 11d = 40
⇒ 11d = 22
⇒ d = 2
∴ a = 9 − 3 × 2 = 3
Thus, the given A.P. is a, a + d, a + 2d …, where a = 3 and d = 2.
Thus, the A.P. is 3, 5, 7, 9 …
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