Question : The average of $n$ numbers is $a$. The first number is increased by 2, the second one is increased by 4, the third one is increased by 8, and so on. The average of the new numbers is:
Option 1: $a+\frac{2^{n–1}–1}{n}$
Option 2: $a+\frac{2(2n–1)}{n}$
Option 3: $a+\frac{2^{n–1}}{n}$
Option 4: $a+\frac{2^{n}–1}{n}$
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Correct Answer: $a+\frac{2(2n–1)}{n}$
Solution : Given: The average of $n$ numbers is $a$. The sum of $n$ numbers is $n×a$. According to the question, Let S be the sum of increased value and $r$ = common ratio = $\frac{4}{2}= 2$ S = 2 + 4 + 8 + 16 .... upto $n$ terms. It is a geometric series. S = $\frac{a(r^{n}–1)}{r–1}=\frac{2(2^{n}–1)}{2–1}= 2(2^{n}–1)$ Now, the sum of old numbers and increased value is $(n×a)+2(2^{n}–1)$. So, the required average = $\frac{na+2(2^{n}–1)}{n}=a+\frac{2(2^{n}–1)}{n}$ Hence, the correct answer is $a+\frac{2(2^{n}–1)}{n}$.
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Question : The average of 5 consecutive integers starting with 'm' is n. What is the average of 6 consecutive integers starting with (m + 2)?
Option 1: $\frac{2n+5}{2}$
Option 2: $(n+2)$
Option 3: $(n+3)$
Option 4: $\frac{2n+9}{2}$
Question : The value of $5–\frac{8+2\sqrt{15}}{4}–\frac{1}{8+2\sqrt{15}}$ is equal to:
Option 1: $\frac{1}{4}$
Option 2: $1$
Option 3: $\frac{2}{3}$
Option 4: $\frac{1}{2}$
Question : The third proportional of $a$ and $\frac{b^4}{4a}$ is:
Option 1: $\frac{b^8}{16 a^2}$
Option 2: $\frac{b^8}{8 a^2}$
Option 3: $\frac{b^8}{8 a^3}$
Option 4: $\frac{b^8}{16 a^3}$
Question : Of the three numbers, the first is 4 times the second and 3 times the third. If the average of all three numbers is 95, what is the third number?
Option 1: 76
Option 2: 60
Option 3: 130
Option 4: 57
Question : The sum of three numbers is 252. If the first number is thrice the second and third number is $\frac{2}{3}$rd of the first, then the second number is:
Option 1: 41
Option 2: 21
Option 3: 42
Option 4: 84
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