JEE Main 2020 — Sequences And Series Question with Solution
From: JEE Main 2020 (Online) 6th September Evening Slot
Question
The common difference of the A.P.
b1, b2, … , bm is 2 more than the common
difference of A.P. a1, a2, …, an. If
a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :
b1, b2, … , bm is 2 more than the common
difference of A.P. a1, a2, …, an. If
a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :
Choose an option
Show full solutionCorrect option: D
Correct answer
D-81
Step-by-step explanation
Let common difference of series
a1 , a2 , a3 ,..., an be d.
a40 = a1 + 39d == –159 ...(i)
and a100 = a1 + 99d = –399 ...(ii)
From eqn. (ii) and (i)
d = –4 and a1 = –3.
The common difference of the A.P.
b1, b2, … , bm is 2 more than the common
difference of A.P. a1, a2, …, an.
Common difference of b1 , b2 , b3 , ..., be (–2).
b100 = a70
b1 + 99(–2) = (–3) + 69(–4)
b1 = 198 – 279
b1 = – 81
a1 , a2 , a3 ,..., an be d.
a40 = a1 + 39d == –159 ...(i)
and a100 = a1 + 99d = –399 ...(ii)
From eqn. (ii) and (i)
d = –4 and a1 = –3.
The common difference of the A.P.
b1, b2, … , bm is 2 more than the common
difference of A.P. a1, a2, …, an.
Common difference of b1 , b2 , b3 , ..., be (–2).
b100 = a70
b1 + 99(–2) = (–3) + 69(–4)
b1 = 198 – 279
b1 = – 81
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