1. The sum of four consecutive numbers in an A.P. is 36 and the ratio of products of first and last terms to the product of the middle terms is 45:77. Using the above information, answer the following questions (i) First term of the A.P. can be
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Let the 4 consecutive numbers in AP be
(a−3d),(a−d),(a+d),(a+3d)
According to the question,
a−3d+a−d+a+d+a+3d=32
4a=32
a=8
Now,
(a−d)(a+d)
(a−3d)(a+3d)
=
15
7
15(a
2
−9d
2
)=7(a
2
−d
2
)
15a
2
−135d
2
=7a
2
−7d
2
8a
2
=128d
2
Putting the value of a, we get,
d
2
=4
d=±2
So, the four consecutive numbers are 2,6,10,14 or 14,10,6,2.
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