S 1- S 2- S 3- S n is the sum of an infinite GP with first terms as 1-2-3 - and common ratios as 1-2- 1-3-1- n -1 then find S 1- S 2- S 3- S nA- 1-2 n2B- 1-2 nn-3C- nD- nn-1

The correct option is B S_1=2 S_2=3 S_1+S_2=5 Here n=2 Go from answer options n(n+1)=2.3 =6 1/2n^2=1/2× 4=2 1/2n(n+3)=1/2(2)(5)=5 n=2 n^2=4 ...

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Quantitative Aptitude

Progressions

S 1, S 2, S 3...

Question

$S_{1}, S_{2}, S_{3} . . . . S_{n}$ is the sum of an infinite GP with first terms as 1,2,3....n and common ratios as $\frac{1}{2}, \frac{1}{3} . . . . \frac{1}{n + 1}$ then find $S_{1} + S_{2} + S_{3} . . . . + S_{n}$

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Solution

The correct option is C

$S_{1} = 2$
$S_{2} = 3$
$S_{1} + S_{2} = 5$
Here n=2
Go from answer options
n(n+1)=2.3 =6
$\frac{1}{2} n^{2} = \frac{1}{2} \times 4 = 2$

$\frac{1}{2} n (n + 3) = \frac{1}{2} (2) (5) = 5$
n=2
$n^{2} = 4$

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S1,S2,S3....Sn is the sum of an infinite GP with first terms as 1,2,3....n and common ratios as 12,13....1n+1 then find S1+S2+S3....+Sn

The correct option is C S1=2 S2=3 S1+S2=5 Here n=2 Go from answer options n(n+1)=2.3 =6 12n2=12×4=2 12n(n+3)=12(2)(5)=5 n=2 n2=4

$S_{1}, S_{2}, S_{3} . . . . S_{n}$ is the sum of an infinite GP with first terms as 1,2,3....n and common ratios as $\frac{1}{2}, \frac{1}{3} . . . . \frac{1}{n + 1}$ then find $S_{1} + S_{2} + S_{3} . . . . + S_{n}$

The correct option is C

$S_{1} = 2$
$S_{2} = 3$
$S_{1} + S_{2} = 5$
Here n=2
Go from answer options
n(n+1)=2.3 =6
$\frac{1}{2} n^{2} = \frac{1}{2} \times 4 = 2$

$\frac{1}{2} n (n + 3) = \frac{1}{2} (2) (5) = 5$
n=2
$n^{2} = 4$