Question:
The common difference of the A.P. \(b_1,b_2,\ldots,b_m\) is 2 more than the common difference of A.P. \(a_1,a_2,\ldots,a_n\). If \(a_{40}=-159\), \(a_{100}=-399\) and \(b_{100}=a_{70}\), then \(b_1\) is equal to :
The common difference of the A.P. \(b_1,b_2,\ldots,b_m\) is 2 more than the common difference of A.P. \(a_1,a_2,\ldots,a_n\). If \(a_{40}=-159\), \(a_{100}=-399\) and \(b_{100}=a_{70}\), then \(b_1\) is equal to :
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Find the common difference of the a-sequence from \(a_{40}\) and \(a_{100}\), then use \(b_{100}=a_{70}\).
Updated On: Nov 5, 2026
- \(127\)
- \(81\)
- \(-127\)
- \(-81\)
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The Correct Option is D
Solution and Explanation
For the a-sequence: \(a_{100}-a_{40}=60d \Rightarrow -399+159=-240=60d \Rightarrow d=-4\).
\(a_{70}=a_{40}+30d=-159-120=-279\).
Common difference of b: \(D=d+2=-2\).
\(b_{100}=b_1+99D \Rightarrow -279=b_1-198 \Rightarrow b_1=-81\).
Answer: (4) \(-81\)
\(a_{70}=a_{40}+30d=-159-120=-279\).
Common difference of b: \(D=d+2=-2\).
\(b_{100}=b_1+99D \Rightarrow -279=b_1-198 \Rightarrow b_1=-81\).
Answer: (4) \(-81\)
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