SSC CGL 2022 PYQ — Question 68

The correct answer is Option 3 i.e. Only I.

For statement I:

(100² - 99²) + (98² - 97²) + (96² - 95²) + (94² - 93²) + ........... + (22² - 21²)

⇒ (100 + 99)(100 - 99) + (98 + 97)(98 - 97) + .......+(22 + 21)(22 - 21)

⇒ 100 + 99 + 98 + 97 + .....+ 22 + 21

Now, sum of first n terms = n(n+1)/2

So, sum of 1 + 2 + 3 + 4 + ....+ 20 = 20 × 21/2 = 210

Also, sum of 1 + 2 + 3 + 4 + ......+ 97 + 98 + 99 + 100 = 100 × 101/2 = 5050

Now thw required sum = 5050 - 210 = 4840.

So, statement I is correct.

for statement II:

$(k^2+{{1} \over k^2})(k-{{1} \over k})(k^4+{{1} \over k^4})(k+{{1} \over k})(k^4-{{1} \over k^4})$

⇒ $({k^2 + {1 \over k^2}})(k^2 - {1 \over k^2})(k^8 - {1 \over k^8})$

⇒ $(k^4 - {1\over k^4})(k^8 - {1 \over k^8})$ which is not equal to ^{$(k^{16}-{{1} \over k^{16}})$}

^{So statement II is incorrect.}

^{Hence statement I is correct.}