Question

The difference between the marks scored by any two consecutive rankers is same. If the sum of marks scored by bottom 'a' rankers is 5a2 + 2a, the marks scored by 8th ranker from the bottom is - a. 83 b. 47 C. 80 d 77​

Answer 1

Given :- The difference between the marks scored by any two consecutive rankers is same. If the sum of marks scored by bottom 'a' rankers is 5a² + 2a, the marks scored by 8th ranker from the bottom is :-

a. 83

b. 47

C. 80

d 77

Solution :-

given that, The difference between the marks scored by any two consecutive rankers is same . Then , we can conclude that, marks scored by all rankers are in AP .

so,

→ S(a) = 5a² + 2a

→ S(1) = 5(1)² + 2*1 = 5 + 2 = 7 = first term

and,

→ S(2) = 5(2)² + 2*2 = 5*4 + 4 = 24

then,

→ second term = 24 - 7 = 17 .

therefore,

→ common difference = 17 - 7 = 10

hence,

→ T(8) = a + (n - 1)d

→ T(8) = 7 + 7 * 10

→ T(8) = 7 + 70

→ T(8) = 77 (d) (Ans.)

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Answer 2

Given : The difference between the marks scored by any two consecutive rankers is same.

the sum of marks scored by bottom 'a' rankers is 5a² + 2a,

To Find :

the marks scored by 8th ranker from the bottom is -

a. 83 b. 47 C. 80 d 77

Solution:

The difference between the marks scored by any two consecutive rankers is same.

Hence Sequence is AP

as aₙ = Sₙ - Sₙ₋₁

marks scored by 8th ranker from the bottom = a₈

a₈ = S₈ - S₇

a₈ =(5(8)² + 2(8))  - ( 5(7)² + 2(7))

=> a₈ =(320 +16)  - ( 245 +14)

=> a₈ =(336)  - ( 259)

=>a₈ =  (336)  - ( 259)

=> a₈ =  77

marks scored by 8th ranker from the bottom =  77

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