TS ICET 2017 Question Paper with Solutions for online practice

Question : 128 of 200

Marks: +1, -0

S_1,S_2

and

S_3

denote the sum ot the first n, 2n and 3n terms ot an Arithmetic Progression, then

S_1/{S_2-S_1}

Solution:

{S}_{1}={S}_{{n}}={n}/{2}[2a+(n-1)d]

....(i)

{S}_{2}={S}_{2{n}}=\;\;{2n}/{2}[2a+(2n-1)d]

....(ii)

S_{3}=S_{3n}=\;\;{3n}/{2}[2a+(3n-1)d]

...(iii)

{S}_{2}-{S}_{1}=\;\;{n}/{2}[(4a+4nd-2d)-(2a+nd-d]

=\;\;{n}/{2}[2a+3nd-d]

or,

S_{2}-S_{1}=\;\;{n}/{2}[2a+(3n-1)d]

.....(iv)
From (iii) and (iv),

\;\;{{S}_{3}}/{{S}_{2}-{S}_{1}}=3

Go to Question:

Prev Question

Next Question