SSC CGL Tier1 Quantitative Aptitude Number system Test 2

The sum of two odd number is even. So, a + b is even.

This is an A.P. in which $ a $ = 6, $ d $ = 6 and S_n = 1800

Then,$ \dfrac{n}{2} $[2$ a $ +$\left ( n - 1\right)$$ d $] = 1800

$\Rightarrow$ $ \dfrac{n}{2} $[2 x 6 + $\left( n - 1\right)$ x 6] = 1800

$\Rightarrow$ 3$ n $ $\left(n + 1\right)$ = 1800

$\Rightarrow n \left(n + 1\right)$ = 600

$\Rightarrow n $² + $ n $ - 600 = 0

$\Rightarrow n $² + 25$ n $ - 24$ n $ - 600 = 0

$\Rightarrow n $$\left( n + 25\right)$ - 24$\left( n + 25\right)$ = 0

$\Rightarrow$ $\left( n + 25\right)$$\left( n - 24\right)$ = 0

$\Rightarrow n $ = 24

Number of terms = 24.

Sum of the squares of first n even numbers = $\dfrac{2n(n + 1)(2n + 1)}{3}$
Sum of squares of first 45 even numbers = $\dfrac{2 × 45 (45 + 1)(2 × 45 + 1)}{3}$
= $\dfrac{90 × 46 × 91}{3}$
= 125580

Sum of cubes of first n natural numbers = $[\dfrac{n(n + 1)}{2}]^{2}$
Plug in each of the options to find the answer which is = 4356.
n = 10; Sum of cubes of first 10 numbers = $[\dfrac{10 × 11}{2}]^{2}$ = 3025.
n = 11; Sum of cubes of first 11 numbers = $[\dfrac{11 × 12}{2}]^{2}$ = 4356.

Sum of cubes of first n natural numbers = $[\dfrac{n(n + 1)}{2}]^{2}$
Here the sum of cubes is a perfect square of some number. Figure out if any of the options is not a perfect square. That will be the answer.
1 is a perfect square [Eliminate]
23409 is a perfect square [Eliminate]
8521 is not a perfect square [= Answer]

Sum of first n even numbers = $n^{2}$ + n
Sum of first n odd numbers = $n^{2}$
So, sum of first n even numbers = Sum of first n odd number + n.

Sum of squares of n numbers is given by the formula $\dfrac{n × (n + 1) (2n + 1)}{6}$
Plug in each of the options and find the sum which = 1240
n = 12; S12 = [12 × (12+1) (2×12 + 1)]/6 = 650
Since n = 12 gives too low a value than 1250, skip n = 13 and try n = 15.
= [15 × 16 × 31] /6
= 1240

Sum of squares of first n natural numbers = $\dfrac{n × (n + 1) (2n + 1)}{6}$
Try estimating the options by picking values for n:
n = 10; [n(n+1)(2n+1)]/6 = 385
n = 11; [n(n+1)(2n+1)]/6 = 506
Notice that the option (a) 420 falls in between the two values 385 (n = 10) and 506 (n = 11). This means, it is not a sum of squares of first n natural numbers. Hence this is the answer.

Sum of squares of first n natural numbers = $\dfrac{n × (n + 1) (2n + 1)}{6}$
Squares of sum of first n natural numbers = $\dfrac{n × (n + 1)}{2} × \dfrac{n × (n + 1)}{2}$
Now the ratio is $\dfrac{n × (n + 1) (2n + 1)}{6}$ : $\dfrac{n × (n + 1)}{2} × \dfrac{n × (n + 1)}{2}$
Plug in the values of each option and check the ratio.
(a) n = 15; (15 × 16 × 31)/6 : (15 × 16 × 15 × 16)/4 = 40 × 31 : 120 × 120 which not in the ratio of 17 : 325. [Eliminate]
(b) n = 25; (25 × 26 × 51)/6 : (25 × 26 × 25 × 26)/4 = 25 × 13 × 17 : 325 × 325 which is in the ratio 17 : 325 [ = Answer]
No need to check for the remaining options after this.

Sum of square of n numbers is found using the formula $\dfrac{n × (n + 1) (2n + 1)}{6}$
Sum of 1st 15 numbers = 1240
Sum of squares from 16 to 30 = Sum of squares of 1st 30 +ve integers - Sum of squares of 1st 15 +ve integers.
= $\dfrac{30 × 31 × 61}{6}$ - 1240
= 9455 – 1240
= 8215