Question

If a,b,c are in GP and a,p,q are in AP such that 2a, b+p, c+q are in GP then the common difference of AP is …….. (This question has multiple correct options)
A. $\sqrt {2a} $
B. $(\sqrt {2a} +1)(a-b)$
C. $\sqrt 2(a-b)$
D. $(\sqrt {2a} -1)(b-a)$

Answer 1

Hint: We know for AP and GP series that If $a,b,c$ are in GP series then we can write ${b^2} = ac$ and If $a,p,q$ are in AP series then we can write $2p = a + q$. Similarly $2a,b + p,c + q$ are in GP series, so we can write ${(b + p)^2} = 2a(c + q)$. Simplify the above three equations and solve the equation by use of quadratic formula $b = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$ and find the difference of AP series as $d = p - a = q - p$.

Complete step by step solution:Here the terms $a,b,c$ are in GP series. For GP series the ratio between two consecutive terms will be the same. We can write as $r = \dfrac{b}{a} = \dfrac{c}{b}$. So, we can write as ${b^2} = ac$.
Terms $a,p,q$ are in AP series. For AP series the difference between two consecutive terms will be the same. We can write as $d = p - a = q - p$. So, we can write $2p = a + q$.
Similarly $2a,b + p,c + q$terms are in GP series. So, we can write as ${(b + p)^2} = 2a(c + q)$.
So, we have three equations:
${b^2} = ac$…..(1)
$2p = a + q$…..(2)
${(b + p)^2} = 2a(c + q)$…..(3)
Simplifying the equation (3),
${(b + p)^2} = 2a(c + q)$
So, \[{b^2} + 2bp + {p^2} = 2ac + 2aq\]
Putting the value of equation (1) in the above equation,
 \[{b^2} + 2bp + {p^2} = 2{b^2} + 2aq\]
Arranging the terms,
\[0 = 2{b^2} - {b^2} + 2aq - 2bp - {p^2}\]
So, \[{b^2} + 2aq - 2bp - {p^2} = 0\]
From equation (2) $2p = a + q$, we can write $p = \dfrac{{a + q}}{2}$
Putting this equation in \[{b^2} + 2aq - 2bp - {p^2} = 0\]
So, \[{b^2} + 2aq - 2b(\dfrac{{a + q}}{2}) - {(\dfrac{{a + q}}{2})^2} = 0\]
Simplifying, \[{b^2} + 2aq - b(a + q) - (\dfrac{{{a^2} + 2aq + {q^2}}}{4}) = 0\]
So, \[{b^2} + 2aq - ab - bq - \dfrac{{{a^2}}}{4} - \dfrac{{2aq}}{4} - \dfrac{{{q^2}}}{4} = 0\]
So, \[{b^2} + 2aq - ab - bq - \dfrac{{{a^2}}}{4} - \dfrac{{aq}}{2} - \dfrac{{{q^2}}}{4} = 0\]
Simplifying, \[{b^2} - (a + q)b - \dfrac{{{a^2}}}{4} + \dfrac{{3aq}}{2} - \dfrac{{{q^2}}}{4} = 0\]
So, \[{b^2} - (a + q)b - (\dfrac{{{a^2}}}{4} - \dfrac{{3aq}}{2} + \dfrac{{{q^2}}}{4}) = 0\]
Above equation is a quadratic equation and root of the quadratic equation of form $A{b^2} + Bb + C = 0$ can be obtained by using formula $b = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$.
Comparing the equation with standard form we can say $A = 1$, $B = - (a + q)$ and $C = - (\dfrac{{{a^2}}}{4} - \dfrac{{3aq}}{2} + \dfrac{{{q^2}}}{4})$.
So, putting the values of A,B and C in $b = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
$b = \dfrac{{ - ( - (a + q)) \pm \sqrt {{{(a + q)}^2} - 4(1)( - (\dfrac{{{a^2}}}{4} - \dfrac{{3aq}}{2} + \dfrac{{{q^2}}}{4}))} }}{{2(1)}}$
So, $b = \dfrac{{(a + q) \pm \sqrt {{{(a + q)}^2} + 4(\dfrac{{{a^2}}}{4} - \dfrac{{3aq}}{2} + \dfrac{{{q^2}}}{4})} }}{2}$
Simplifying, \[b = \dfrac{{(a + q) \pm \sqrt {{{(a + q)}^2} + 4(\dfrac{{{a^2}}}{4}) - 4(\dfrac{{3aq}}{2}) + 4(\dfrac{{{q^2}}}{4})} }}{2}\]\
So, \[b = \dfrac{{(a + q) \pm \sqrt {{{(a + q)}^2} + {a^2} - 6aq + {q^2}} }}{2}\]
Using ${(a + b)^2} = {a^2} + 2ab + {b^2}$ formula, \[b = \dfrac{{(a + q) \pm \sqrt {{a^2} + 2aq + {q^2} + {a^2} - 6aq + {q^2}} }}{2}\]
Simplifying, \[b = \dfrac{{(a + q) \pm \sqrt {2{a^2} - 4aq + 2{q^2}} }}{2}\]
So, \[b = \dfrac{{(a + q) \pm \sqrt {2({a^2} - 2aq + {q^2})} }}{2}\]
So, \[b = \dfrac{{(a + q) \pm \sqrt {2{{(a - q)}^2}} }}{2}\]
Simplifying, \[b = \dfrac{{(a + q) \pm \sqrt 2 (a - q)}}{2}\]
So, root \[b = \dfrac{{(a + q) + \sqrt 2 (a - q)}}{2}\] and \[b = \dfrac{{(a + q) - \sqrt 2 (a - q)}}{2}\]
So, \[2b = (a + q) + \sqrt 2 (a - q)\] and \[2b = (a + q) - \sqrt 2 (a - q)\]
As $a,p,q$ are in AP series, we can write that $q = a + 2d$
So, \[2b = (a + a + 2d) + \sqrt 2 (a - (a + 2d))\] and \[2b = (a + a + 2d) - \sqrt 2 (a - (a + 2d))\]
Simplifying, \[2b = (2a + 2d) + \sqrt 2 ( - 2d))\] and \[2b = (2a + 2d) - \sqrt 2 ( - 2d))\]
Dividing both side by 2, \[b = (a + d) - \sqrt 2 d\] and \[b = (a + d) + \sqrt 2 d\]
Taking the terms common, \[b = a + (1 - \sqrt 2 )d\]and \[b = a + (1 + \sqrt 2 )d\]
So, \[d = \dfrac{{b - a}}{{(1 - \sqrt 2 )}}\] and \[d = \dfrac{{b - a}}{{(1 + \sqrt 2 )}}\]
Simplifying, \[d = \dfrac{{b - a}}{{(1 - \sqrt 2 )}} \cdot \dfrac{{1 + \sqrt 2 }}{{(1 + \sqrt 2 )}}\] and \[d = \dfrac{{b - a}}{{(1 + \sqrt 2 )}} \cdot \dfrac{{1 - \sqrt 2 }}{{(1 - \sqrt 2 )}}\]
So, \[d = \dfrac{{(b - a) \cdot (1 + \sqrt 2 )}}{{{{(1)}^2} - {{(\sqrt 2 )}^2}}}\] and \[d = \dfrac{{(b - a) \cdot (1 - \sqrt 2 )}}{{{{(1)}^2} - {{(\sqrt 2 )}^2}}}\]
So, \[d = \dfrac{{(b - a) \cdot (1 + \sqrt 2 )}}{{1 - 2}}\] and \[d = \dfrac{{(b - a) \cdot (1 - \sqrt 2 )}}{{1 - 2}}\]
Simplifying, \[d = \dfrac{{(\sqrt 2 + 1) \cdot (b - a)}}{{ - 1}}\] and \[d = \dfrac{{(1 - \sqrt 2 ) \cdot (b - a)}}{{ - 1}}\]
So, \[d = (\sqrt 2 + 1) \cdot (a - b)\] and \[d = (\sqrt 2 - 1) \cdot (b - a)\]

So, Option (B) and Option (D) both are correct answers.

Note: In the same question as $a,b,c$are in GP series, we can calculate the ratio of the GP series by solving the three equations as stated above in terms of a and q. Also we can calculate the ratio of GP series of terms $2a,b + p,c + q$ by simplifying and solving the three equations.