Question

Find a, so that a, a + 2, a+ 6 are consecutive terms of a GP:

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer:

A. 2

${a, a + 2, a+ 6}$$$are in GP

$\implies{\frac{a+2}{a}}=\frac{a+6}{a+2}$

$(a+2)^2=a(a+6)$

$a^2+4a+4=a^2+6a$

$2a=4$

$a=4/2=2$