For an A. P, $S_{100} = 3 S_{50}$ The value of $S_{150} : S_{50}$ $=$

$8$

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$3$

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$6$

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$10$

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Solution

The correct option is C
$6$

Explanation for correct option.
Step 1: Given relation in Arithmetic Progression
$S_{100} = 3 S_{50}$
Sum of $n$ terms in AP = $S_{n} = \frac{n}{2} [[2 a + (n - 1) d]]$
Where , $S_{n}$ = sum of $n$ terms of an A.P
$a$ = First term if an A.P
$d$ = Common difference between two consecutive terms of an Arithmetic Progression
Step 2: Using formula to drive relation between $a$ and $d$
$S_{100} = 3 S_{50}$
$\begin{array}{l} \frac{100}{2} [2 a + (100 - 1) d] = 3 \times \frac{50}{2} [2 a + (50 - 1) d] \\ \Rightarrow 2 [2 a + (99) d] = 3 [2 a + (49) d] \\ \Rightarrow 4 a + 198 d = 6 a + 147 d \\ \Rightarrow 2 a = 51 d - - - - - - - - - (1) \end{array}$
Step 3: Using the value of a and putting it in the equation to find
$S_{150} : S_{50}$
$\Rightarrow \frac{150}{2} [2 a + (150 - 1) d] : \frac{50}{2} [2 a + (50 - 1) d] \Rightarrow 3 [2 a + (149) d] : [2 a + (49) d] \Rightarrow 3 [51 d + 149 d] : [51 d + 49 d] \Rightarrow 3 [200 d] : [100 d] \Rightarrow 6 : 1 o r 6$

Therefore, the value of $S_{150} : S_{50}$ is $6$ .
Hence, the correct option is $(C)$

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