JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 13)
Let $$a$$, b and c be in G.P. with common ratio r, where $$a$$ $$ \ne $$ 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 :
$$a$$
$${7 \over 3}a$$
5$$a$$
$${2 \over 3}a$$
Explanation
a = a, b = ar and c = ar2
3a, 7b, 15c $$ \to $$ A.P.
14b = 3a + 15c
14(ar) = 3a + 15(ar2)
15r2 – 14r + 3 = 0
$$ \Rightarrow r = {1 \over 3},{3 \over 5}(rejected)$$
Common difference = 7b – 3a
= 7ar – 3a
$$ \Rightarrow $$ $${{7a} \over 3} - 3a = - {2 \over 3}a$$
4th term is $$ \Rightarrow $$ $$15c - {2 \over 3}a = {{15} \over 9}a - {2 \over 3}a = a$$
3a, 7b, 15c $$ \to $$ A.P.
14b = 3a + 15c
14(ar) = 3a + 15(ar2)
15r2 – 14r + 3 = 0
$$ \Rightarrow r = {1 \over 3},{3 \over 5}(rejected)$$
Common difference = 7b – 3a
= 7ar – 3a
$$ \Rightarrow $$ $${{7a} \over 3} - 3a = - {2 \over 3}a$$
4th term is $$ \Rightarrow $$ $$15c - {2 \over 3}a = {{15} \over 9}a - {2 \over 3}a = a$$
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