An a.p has 7 terms, then the 6th term is 9 more than the 3rd term and the product of the first and the last term is is 40, write the sequence.
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Given,
The AP has 7 terms.
6th term = 9 more than the 3rd term
Product of the first and last term = 40
To find,
The AP.
Solution,
The AP will be 2, 5, 8, --- or 20, -17, -14, ---.
According to the question,
The AP has 7 terms.
6th term = 9 more than the 3rd term
a6 = 9+a3
[The formula to find an is a+(n-1)d.]
a+(6-1)d = 9+a+(3-1)d
a+5d = 9+a+2d
a-a+5d = 9+2d
5d = 9+2d
Moving 2d from RHS to LHS,
5d-2d = 9
3d = 9
d = 9/3
d = 3
Product of the first and last term = 40
a. a7 = 40
a. [a+(7-1)d] = 40
a(a+6d) = 40
Putting the value of d,
a(a+18) = 40
a²+18a = 40
a²+18a-40 = 0
Factorising the expression by splitting the middle term:
a²+20a-2a-40 = 0
a(a+20)-2(a+20) = 0
(a+20)(a-2) = 0
a-2 = 0 or a+20 = 0
a = 2 or -20
So, the AP will be:
2, 5, 8, --- (adding the value of d to the first term to get the second term)
or -20, -17, -14, ---
Hence, the possible AP are 2, 5, 8, --- or 20, -17, -14, ---.