Math, asked by meandonlyme, 1 year ago

Q.24 pls.this is urgently needed

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Answers

Answered by Draxillus

hey dear,

refer to given attachment please

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meandonlyme: i am really sorry.but i expected the ans for q 24

Answered by aryan223344177662617

Let the four consecutive terms of that AP be ...
(a-3d),(a-d),(a+d),(a+3d)
Then... Acc. To que.
a-3d + a-d +a +d +a+3d =32
4a = 32
a =8......(1)
Now ... Acc. to que.
(a-3d)(a+3d)/(a-d)(a+d) =7/15

15{a²- (3d)²}=7{a²-d²}
15(a²-9d²)=7a²-7d²
15a²-135d²=7a²-7d²
15a²-135d²-7a²+7d²=0
8a²-128d²=0
8a²=128d²
a²=16d²
a=4d
Put the value of eqn.1
8=4d
d=2
Now ...
a-3d=8-3(2)=2
a-d=8-2=6
a+d=8+2=10
a+3d=8+3(2)=14

Hope this will help...

meandonlyme: hello.cant the consecutive terms be a-2d,a-d,a,a+d

meandonlyme: ??

aryan223344177662617: Is my ans. Correct??

meandonlyme: its right

meandonlyme: can u explain why u held the terms as you stated above

meandonlyme: ??

aryan223344177662617: No..if we take these consecutive numbers as above...it will be more easy to solve ...as "d" will be cancelled in the first case...

aryan223344177662617: Ok??

meandonlyme: oh thanks alot

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