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 :
Solution
Let common difference of series
<br>a<sub>1</sub>
, a<sub>2</sub>
, a<sub>3</sub>
,..., a<sub>n</sub>
be d.
<br><br>$\because$ a<sub>40</sub> = a<sub>1</sub> + 39d == –159 ...(i)
<br><br>and a<sub>100</sub> = a<sub>1</sub> + 99d = –399 ...(ii)
<br><br>From eqn. (ii) and (i)
<br>d = –4 and a<sub>1</sub>
= –3.
<br><br>The common difference of the A.P. <br>b<sub>1</sub>, b<sub>2</sub>, … , b<sub>m</sub>
is 2 more than the common<br> difference of A.P. a<sub>1</sub>, a<sub>2</sub>, …, a<sub>n</sub>.
<br><br>$\therefore$ Common difference of b<sub>1</sub>
, b<sub>2</sub>
, b<sub>3</sub>
, ..., be (–2).
<br><br>$\because$ b<sub>100</sub> = a<sub>70</sub>
<br><br>$\therefore$ b<sub>1</sub>
+ 99(–2) = (–3) + 69(–4)
<br><br>$\therefore$ b<sub>1</sub>
= 198 – 279
<br><br>$\therefore$ b<sub>1</sub>
= – 81
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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