Question no:11,12,13
In a need
Answers
Answer:
29 numbers are divisible by 3.
Answer:
11) It should be an ap whose first term is 12 and last term is 99
12,15,18,21- - - - - - -99
a=12
l=99
n=?
d=3
l=a+(n-1)d
99=12+(n-1)3
87/3=n-1
29=n-1
n=30
12) Answer:
Step-by-step explanation:
Given the equation
\frac{16}{x}-1=\frac{15}{x+1}\text{, x is not equal to 0 or -1.}
we have to solve the above equation.
\frac{16}{x}-1=\frac{15}{x+1}
\frac{16-x}{x}=\frac{15}{x+1}
Cross multiplying both sides, we get
(16-x)(x+1)=15x
16x+16-x^2-x=15x
16x-15x-x+16=x^2
x^2=16
Taking square root on both sides, we get
x=\pm 4
13) step by step;
let sn denote the sum of the first n terms of AP
therefore; Sn = 3 n^2 +5n
⇒ 3(n^2 - 2n +1)^2 + 5(n-1)
⇒ 3n^2-n-2
Now,
n term of AP, an=sn-(sn-1)
⇒ (3n^2 + 5n)-(3n^2-n-2)
⇒ 6n+2
let 'd' be the common difference of the AP
∴ d=an-(an-1)
⇒ (6n+2)-[6(n-1)+2]
⇒ 6n+2-6(n-1)-2
⇒ 6
∴ the answer is 6.
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