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Yogita Ingle 5 years, 2 months ago
Let the present age of Nuri =x year
And present age of Sonu =y year
Five years ago
Age of Nuri =x–5years
Age of Sony =y–5years
Nuri was thrice as old as Sonu
X−5=3(y–5)
X–5=35–15
X–3y=−15+5
X–3y=−10 ………..(1)
Ten years later,
Age of Nuri =x+10
Age of Sonu =y+10
Nuri will be twice as old as Sonu.
X+10=2(y+10)
X+10=2y+20
X–2y=10 ………..(2)
X–3y=−10 ………..(1)
Subtracting equation (1) from equation (2) we get
Y=20
Plug this value in equation first we get
X−3∗20=−10
X=60–10
X=50
Hence age of Nuri =50 years and age of Sonu =20 years
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Ram Kushwah 5 years, 2 months ago
2x²+5px+7q=0
Let the roots are a and 1/a
Product of roots=a*1/a=1
so 7q/2=1
So q=2/7
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Yogita Ingle 5 years, 2 months ago
Let a and d be the first term and common difference of AP
nth term of AP
an=a+(n−1)d
∴a3=a+(3−1)d=a+2d
a7=a+(7−1)d=a+6d
Given a3+a7=6
∴(a+2d)+(a+6d)=6
⇒2a+8d=6
⇒a+4d=3....(1)
Also given
a3×a7=8
∴(a+2d)(a+6d)=8
⇒(3−4d+2d)(3−4d+6d)=8 [Using (1)]
⇒(3−2d)(3+2d)=8
⇒9−4d2=8
⇒4d2=1
⇒d2=1/4
⇒d=± 1/2
When d=1/2
a=3−4d=3−4×(1/2) =3−2=1
When d=−1/2
a=3−4d=3+4×(1/2) =3+2=5
When a=1 & d=21
S16=216[2×1+(16−1)×21]=8(2+215)=4×19=76
When a=5 & d=−21
S16=216[2×5+(16−1)×(−21)]=8(10−215)=4×5=20
Thus, the sum of first 16 terms of the AP is 76 or 20.

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Meenakshi Kasi 5 years, 2 months ago
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