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Given that : The sum of 4 numbers in A.P. is 40
Let, the numbers are (a-3d), (a-d), (a+d), (a+3d)
Then, according to the question :
(a-3d)+(a-d)+(a+d)+(a+3d) = 40
=> 4a = 40
=> a = 10
Also, given that : The product of their extremes is 91
Again, according to the question :
(a-3d)(a+3d) = 91
=> a²–9d² = 91
=> (10)²–9d² = 91
=> 100 –91 = 9d²
=> 9d² = 9
=> d² = 1
=> d = ±1
when d = 1 then, numbers are :
a-3d = 10-3×1 = 7
a-d = 10-1 = 9
a+d = 10+1 = 11
a+3d = 10+3×1 = 13
When d = 1 then numbers will be 7,9,11 and 13
when d = -1 then, numbers are :
a-3d = 10-3×(-1) = 13
a-d = 10-(-1) = 11
a+d = 10-1 = 9
a+3d = 10+3×(-1) = 7
when d=-1 then, numbers will be 13,11,9 and 7
Let, the numbers are (a-3d), (a-d), (a+d), (a+3d)
Then, according to the question :
(a-3d)+(a-d)+(a+d)+(a+3d) = 40
=> 4a = 40
=> a = 10
Also, given that : The product of their extremes is 91
Again, according to the question :
(a-3d)(a+3d) = 91
=> a²–9d² = 91
=> (10)²–9d² = 91
=> 100 –91 = 9d²
=> 9d² = 9
=> d² = 1
=> d = ±1
when d = 1 then, numbers are :
a-3d = 10-3×1 = 7
a-d = 10-1 = 9
a+d = 10+1 = 11
a+3d = 10+3×1 = 13
When d = 1 then numbers will be 7,9,11 and 13
when d = -1 then, numbers are :
a-3d = 10-3×(-1) = 13
a-d = 10-(-1) = 11
a+d = 10-1 = 9
a+3d = 10+3×(-1) = 7
when d=-1 then, numbers will be 13,11,9 and 7
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