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poojapithadia1p32m2z
@poojapithadia1p32m2z
June 2022
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find 4 consecutive term of a.p whose sum is 66 and product of extremists to product of means is 14:15
Answers & Comments
rohitkumargupta
Verified answer
HELLO DEAR,
assume the terms to be (a-3d), (a-d), (a+d)
and (a+3d)
a-3d +a-d+a+d+a+3d= 66
4a = 66
a= 66/4= 33/2
since product of extremes to the product of
means is 14/15
(a-3d)(a+3d)/(a-d)(a+d) = 14/15
a2 - 9d2/ a2- d2 = 14/15
15 ( a2- 9d2) = 14 ( a2- d2)
15a2- 135d2= 14a2- 14d2
15a2-14a2= 135d2-14d2
a2= 121d2
substituting value of a,
(33/2)2= 121d2
1089/4 = 121d2
d2= 1089/4*121
d= 1.5
so the ap is,
a-3d
= >16.5-4.5= 12
a-d= 16.5- 1.5= 15
a+d = 16.5 + 1.5 = 18
a+3d = 16.5 + 4.5= 21
the ap is = 12,15,18,21
I HOPE ITS HELP YOU DEAR
THANKS
1 votes
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Answers & Comments
Verified answer
HELLO DEAR,assume the terms to be (a-3d), (a-d), (a+d)
and (a+3d)
a-3d +a-d+a+d+a+3d= 66
4a = 66
a= 66/4= 33/2
since product of extremes to the product of
means is 14/15
(a-3d)(a+3d)/(a-d)(a+d) = 14/15
a2 - 9d2/ a2- d2 = 14/15
15 ( a2- 9d2) = 14 ( a2- d2)
15a2- 135d2= 14a2- 14d2
15a2-14a2= 135d2-14d2
a2= 121d2
substituting value of a,
(33/2)2= 121d2
1089/4 = 121d2
d2= 1089/4*121
d= 1.5
so the ap is,
a-3d
= >16.5-4.5= 12
a-d= 16.5- 1.5= 15
a+d = 16.5 + 1.5 = 18
a+3d = 16.5 + 4.5= 21
the ap is = 12,15,18,21
I HOPE ITS HELP YOU DEAR
THANKS