JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

{S_n} = nP + \frac{1}{2}n\left( {n - 1} \right){\cal Q}

, where

{S_n}

denotes the sum of the first

n

terms of an AP, then the common difference is

View answer and explanation

Correct option: D

{\cal Q}

Option D (

{\cal Q}

) is correct because the common difference of an AP can be found by subtracting consecutive terms, which simplifies to

{\cal Q}

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