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Sequences and Series question

2019 · 10 Apr · Shift 2 · Q35

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Sequences and Series
2019 · 10 Apr · Shift 2 · Q35

Sequences and Series question

2019 · 10 Apr · Shift 2 · Q35

JEE MainMathematicsSequences and SeriesMCQ+4 / −1

Let

a

, b and c be in G.P. with common ratio r, where

a e

0 and 0 < r

\le {1 \over 2}

. If 3

a

, 7b and 15c are the first three terms of an A.P., then the 4th term of this A.P. is :

A
$a$
B
${7 \over 3}a$
C
5 $a$
D
${2 \over 3}a$

View written solutionFree

Correct answer: A

Since $a,b,c$ are in G.P. with common ratio $r$ , we have $b=ar, \qquad c=ar^2.$
We are told that $3a,\; 7b,\; 15c$ are the first three terms of an A.P. For three numbers in A.P., the middle term is the average of the other two: $2(7b)=3a+15c.$ Substitute $b=ar$ and $c=ar^2$ : $14ar=3a+15ar^2.$
Since $a\ne 0$ , divide throughout by $a$ : $14r=3+15r^2.$ Rearrange: $15r^2-14r+3=0.$ Factorize: $15r^2-14r+3=(5r-3)(3r-1)=0.$ So, $r=\frac35 \quad \text{or} \quad r=\frac13.$
Given $0<r\le \frac12$ , only $r=\frac13$ is valid.
Now the A.P. has first three terms: $3a,\quad 7b=7\left(\frac a3\right)=\frac{7a}{3},\quad 15c=15\left(a\cdot \frac{1}{9}\right)=\frac{5a}{3}.$ Its common difference is $d=\frac{7a}{3}-3a=\frac{7a-9a}{3}=-\frac{2a}{3}.$
Therefore, the 4th term is $\frac{5a}{3}+\left(-\frac{2a}{3}\right)=\frac{3a}{3}=a.$

So the 4th term of the A.P. is $\boxed{a}.$

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