find four consecutive terms in an ap such that the sum of the middle two terms is 18 and a product of two ends and term is 45
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Answered by
21
the series are 3 7 11 15 or 15 11 7 3
sahilshivhare60:
how???
consider the 4 terms as a-3d,a-d,a+d,a+3d the find values of a and d with given conditions so we get 2 values of d +2,-2 and 1 value of a that is 9 so now put the values in the ap
Answered by
79
Answer:
Step-by-step explanation: let (a-3d),(a-d),(a+d) ,(a+3d) be four consecutive terms in an A.P.
(a-d)+(a+d)=18
2a=18
a=9
(a-3d) (a+3d)=45
(9-3d)(9+3d)=45
9²-(3d)²=45
81-9d²=45
-9d²=45-81
-9d²=(-36)
9d²=36
d²=4
d=2
hence,
(a-3d)= 9-3×2=9-6=3
(a-d)=9-2=7
(a+d)=9+2=11
(a+3d)=9+3×2=9+6=15
therfore terms are 3,7,11,15 (when d positive) and 15,11,7,3(when d is negative)
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