Let a1,a1+d1,a1+2d.................first A.P.
a40=a1+39d=−159.....................(1)
a100=a1+99d=−399....................(2)
d=−4,a1=−3
now
b100=a70
⇒b1+99D=a1+69d
b1+99(−2)=−3+69(−4)
(∵Accoding to question D=d+2)
⇒b1=−81
The common difference of the A.P.b1,b2,....,bm is 2 more than common difference of A.P.a1,a2,.....,an. If a40=−159,a100=−399 and b100=a70, then b1 is equal to :
Held on 6 Sept 2020 · Verified 6 Jul 2026.
81
−127
−81
127
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