Find four consecutive terms in an A.P. such that sum of middle two terms is 18 and product of two end terms is 45
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Let 4 terms are a-3d,a-d,a+d,a+3d
Then a-d+a+d=18
2a=18
a=9
(a-3d)(a+3d)=45
a^2-9d^2=45
81-9d^2+45
d^2=4
d=+ or -2
Substitute in tems
Numbers are 3,7,11,15 or 15,11,7,3
Then a-d+a+d=18
2a=18
a=9
(a-3d)(a+3d)=45
a^2-9d^2=45
81-9d^2+45
d^2=4
d=+ or -2
Substitute in tems
Numbers are 3,7,11,15 or 15,11,7,3
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