Question : Given $\small 2^{2}+4^{2}+6^{2}+......+40^{2}=11480$, then the value of $\small 1^{2}+2^{2}+3^{2}+......+20^{2}$ is:
Option 1: 2870
Option 2: 2868
Option 3: 2867
Option 4: 2869
Correct Answer: 2870
Solution : Given: $2^{2}+4^{2}+6^{2}+......+40^{2}=11480$ $⇒2^{2}(1^{2}+2^{2}+3^{2}+......+20^{2})=11480$ $⇒1^{2}+2^{2}+3^{2}+......+20^{2}=\frac{11480}{4}$ $\therefore1^{2}+2^{2}+3^{2}+......+20^{2}=2870$ Hence, the correct answer is 2870.
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Question : If $\small 1^{2}+2^{2}+3^{2}+......+\ p^{2}=\frac{p(p+1)(2p+1)}{6}$, then $\small 1^{2}+3^{2}+5^{2}+......+17^{2}$ is equal to:
Option 1: 1785
Option 2: 1700
Option 3: 980
Option 4: 969
Question : Directions: Which of the following interchange of numbers (not digits) would make the given equation correct? 16 × 5 ÷ 20 + 6 + 50 = 40
Option 1: 40, 50
Option 2: 6, 40
Option 3: 20, 40
Option 4: 5, 40
Question : If a3 + b3 = 217 and a + b = 7, then the value of ab is:
Option 1: – 6
Option 2: – 1
Option 3: 7
Option 4: 6
Question : If $x^4+y^4=x^2 y^2$, then the value of $x^6+y^6$ is:
Option 1: 2
Option 2: 0
Option 3: 1
Option 4: 3
Question : If $x:y::2:3$ and $2:x::4:8$, then the value of $y$ is:
Option 1: 6
Option 2: 8
Option 3: 4
Option 4: 12
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