Let an be the nth term of an A.P. If ∑r=1100 a2r=α and ∑r=1100 a2r−1=β, then the common difference of the A.P. is:

A
$α - β$
B
$\frac{1}{100} (α - β)$
C
$β - α$
D
$\frac{1}{200} (α - β)$

Solution:

Let $d$ the common difference of the A.P., then
$\begin{array}{l} α - β = \sum_{r = 1}^{100} (a_{2 r} - a_{2 r - 1}) = \sum_{r = 1}^{100} d = 100 d \\ \Rightarrow d = \frac{1}{100} (α - β) \end{array}$

Related content

Test your English Vocabulary

CUET Exam Dates 2024 – Application Form, Fees, Eligibility

CBSE Class 12 IP Answer Key 2024,Informatics Practices Paper Solution For SET 1, 2, 3, 4

CUET UG Cut Off 2024, Category, Universities and Colleges Wise Expected Cut Off

Modal Verbs

Helping Verbs

Letter To Your Friend About Your School Trip

Action Verbs

CUET 2024 – List of Colleges and Participating Universities Accepting CUET Exam Score

SRMJEEE Online Test Series – Practice Papers