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JEE Mains · Maths · STD 11 - 8. sequence and series

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

  1. A \(-127\)
  2. B \(-81\)
  3. C \(81\)
  4. D \(127\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-81\)

Step-by-step Solution

Detailed explanation

\(a_{1}, a_{2}, \ldots, a_{n} \rightarrow(C D=d)\) \(b _{1}, b _{2}, \ldots, b _{ m } \rightarrow( CD = d +2)\) \(a_{40}=a+39 d=-159\) \(a_{100}=a+99 d=-399\) Subtract : \(60 d =-240 \Rightarrow d =-4\) using equation (1) \(a+39(-4)=-159\) \(a=156-159=-3\)…
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