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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 o...

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. 2a\sqrt {2a}
B. (2a+1)(a−b)(\sqrt {2a} +1)(a-b)
C. 2(a−b)\sqrt 2(a-b)
D. (2a−1)(b−a)(\sqrt {2a} -1)(b-a)

Explanation

Solution

We know for AP and GP series that If a,b,ca,b,c are in GP series then we can write b2=ac{b^2} = ac and If a,p,qa,p,q are in AP series then we can write 2p=a+q2p = a + q. Similarly 2a,b+p,c+q2a,b + p,c + q are in GP series, so we can write (b+p)2=2a(c+q){(b + p)^2} = 2a(c + q). Simplify the above three equations and solve the equation by use of quadratic formula b=−B±B2−4AC2Ab = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}} and find the difference of AP series as d=p−a=q−pd = p - a = q - p.

Complete step by step solution: Here the terms a,b,ca,b,c are in GP series. For GP series the ratio between two consecutive terms will be the same. We can write as r=ba=cbr = \dfrac{b}{a} = \dfrac{c}{b}. So, we can write as b2=ac{b^2} = ac.
Terms a,p,qa,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−pd = p - a = q - p. So, we can write 2p=a+q2p = a + q.
Similarly 2a,b+p,c+q2a,b + p,c + qterms are in GP series. So, we can write as (b+p)2=2a(c+q){(b + p)^2} = 2a(c + q).
So, we have three equations:
b2=ac{b^2} = ac…..(1)
2p=a+q2p = a + q…..(2)
(b+p)2=2a(c+q){(b + p)^2} = 2a(c + q)…..(3)
Simplifying the equation (3),
(b+p)2=2a(c+q){(b + p)^2} = 2a(c + q)
So, b2+2bp+p2=2ac+2aq{b^2} + 2bp + {p^2} = 2ac + 2aq
Putting the value of equation (1) in the above equation,
b2+2bp+p2=2b2+2aq{b^2} + 2bp + {p^2} = 2{b^2} + 2aq
Arranging the terms,
0=2b2−b2+2aq−2bp−p20 = 2{b^2} - {b^2} + 2aq - 2bp - {p^2}
So, b2+2aq−2bp−p2=0{b^2} + 2aq - 2bp - {p^2} = 0
From equation (2) 2p=a+q2p = a + q, we can write p=a+q2p = \dfrac{{a + q}}{2}
Putting this equation in b2+2aq−2bp−p2=0{b^2} + 2aq - 2bp - {p^2} = 0
So, b2+2aq−2b(a+q2)−(a+q2)2=0{b^2} + 2aq - 2b(\dfrac{{a + q}}{2}) - {(\dfrac{{a + q}}{2})^2} = 0
Simplifying, b2+2aq−b(a+q)−(a2+2aq+q24)=0{b^2} + 2aq - b(a + q) - (\dfrac{{{a^2} + 2aq + {q^2}}}{4}) = 0
So, b2+2aq−ab−bq−a24−2aq4−q24=0{b^2} + 2aq - ab - bq - \dfrac{{{a^2}}}{4} - \dfrac{{2aq}}{4} - \dfrac{{{q^2}}}{4} = 0
So, b2+2aq−ab−bq−a24−aq2−q24=0{b^2} + 2aq - ab - bq - \dfrac{{{a^2}}}{4} - \dfrac{{aq}}{2} - \dfrac{{{q^2}}}{4} = 0
Simplifying, b2−(a+q)b−a24+3aq2−q24=0{b^2} - (a + q)b - \dfrac{{{a^2}}}{4} + \dfrac{{3aq}}{2} - \dfrac{{{q^2}}}{4} = 0
So, b2−(a+q)b−(a24−3aq2+q24)=0{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 Ab2+Bb+C=0A{b^2} + Bb + C = 0 can be obtained by using formula b=−B±B2−4AC2Ab = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}.
Comparing the equation with standard form we can say A=1A = 1, B=−(a+q)B = - (a + q) and C=−(a24−3aq2+q24)C = - (\dfrac{{{a^2}}}{4} - \dfrac{{3aq}}{2} + \dfrac{{{q^2}}}{4}).
So, putting the values of A,B and C in b=−B±B2−4AC2Ab = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}
b=−(−(a+q))±(a+q)2−4(1)(−(a24−3aq2+q24))2(1)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=(a+q)±(a+q)2+4(a24−3aq2+q24)2b = \dfrac{{(a + q) \pm \sqrt {{{(a + q)}^2} + 4(\dfrac{{{a^2}}}{4} - \dfrac{{3aq}}{2} + \dfrac{{{q^2}}}{4})} }}{2}
Simplifying, b=(a+q)±(a+q)2+4(a24)−4(3aq2)+4(q24)2b = \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=(a+q)±(a+q)2+a2−6aq+q22b = \dfrac{{(a + q) \pm \sqrt {{{(a + q)}^2} + {a^2} - 6aq + {q^2}} }}{2}
Using (a+b)2=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2} formula, b=(a+q)±a2+2aq+q2+a2−6aq+q22b = \dfrac{{(a + q) \pm \sqrt {{a^2} + 2aq + {q^2} + {a^2} - 6aq + {q^2}} }}{2}
Simplifying, b=(a+q)±2a2−4aq+2q22b = \dfrac{{(a + q) \pm \sqrt {2{a^2} - 4aq + 2{q^2}} }}{2}
So, b=(a+q)±2(a2−2aq+q2)2b = \dfrac{{(a + q) \pm \sqrt {2({a^2} - 2aq + {q^2})} }}{2}
So, b=(a+q)±2(a−q)22b = \dfrac{{(a + q) \pm \sqrt {2{{(a - q)}^2}} }}{2}
Simplifying, b=(a+q)±2(a−q)2b = \dfrac{{(a + q) \pm \sqrt 2 (a - q)}}{2}
So, root b=(a+q)+2(a−q)2b = \dfrac{{(a + q) + \sqrt 2 (a - q)}}{2} and b=(a+q)−2(a−q)2b = \dfrac{{(a + q) - \sqrt 2 (a - q)}}{2}
So, 2b=(a+q)+2(a−q)2b = (a + q) + \sqrt 2 (a - q) and 2b=(a+q)−2(a−q)2b = (a + q) - \sqrt 2 (a - q)
As a,p,qa,p,q are in AP series, we can write that q=a+2dq = a + 2d
So, 2b=(a+a+2d)+2(a−(a+2d))2b = (a + a + 2d) + \sqrt 2 (a - (a + 2d)) and 2b=(a+a+2d)−2(a−(a+2d))2b = (a + a + 2d) - \sqrt 2 (a - (a + 2d))
Simplifying, 2b=(2a+2d)+2(−2d))2b = (2a + 2d) + \sqrt 2 ( - 2d)) and 2b=(2a+2d)−2(−2d))2b = (2a + 2d) - \sqrt 2 ( - 2d))
Dividing both side by 2, b=(a+d)−2db = (a + d) - \sqrt 2 d and b=(a+d)+2db = (a + d) + \sqrt 2 d
Taking the terms common, b=a+(1−2)db = a + (1 - \sqrt 2 )dand b=a+(1+2)db = a + (1 + \sqrt 2 )d
So, d=b−a(1−2)d = \dfrac{{b - a}}{{(1 - \sqrt 2 )}} and d=b−a(1+2)d = \dfrac{{b - a}}{{(1 + \sqrt 2 )}}
Simplifying, d=b−a(1−2)⋅1+2(1+2)d = \dfrac{{b - a}}{{(1 - \sqrt 2 )}} \cdot \dfrac{{1 + \sqrt 2 }}{{(1 + \sqrt 2 )}} and d=b−a(1+2)⋅1−2(1−2)d = \dfrac{{b - a}}{{(1 + \sqrt 2 )}} \cdot \dfrac{{1 - \sqrt 2 }}{{(1 - \sqrt 2 )}}
So, d=(b−a)⋅(1+2)(1)2−(2)2d = \dfrac{{(b - a) \cdot (1 + \sqrt 2 )}}{{{{(1)}^2} - {{(\sqrt 2 )}^2}}} and d=(b−a)⋅(1−2)(1)2−(2)2d = \dfrac{{(b - a) \cdot (1 - \sqrt 2 )}}{{{{(1)}^2} - {{(\sqrt 2 )}^2}}}
So, d=(b−a)⋅(1+2)1−2d = \dfrac{{(b - a) \cdot (1 + \sqrt 2 )}}{{1 - 2}} and d=(b−a)⋅(1−2)1−2d = \dfrac{{(b - a) \cdot (1 - \sqrt 2 )}}{{1 - 2}}
Simplifying, d=(2+1)⋅(b−a)−1d = \dfrac{{(\sqrt 2 + 1) \cdot (b - a)}}{{ - 1}} and d=(1−2)⋅(b−a)−1d = \dfrac{{(1 - \sqrt 2 ) \cdot (b - a)}}{{ - 1}}
So, d=(2+1)⋅(a−b)d = (\sqrt 2 + 1) \cdot (a - b) and d=(2−1)⋅(b−a)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,ca,b,care 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+q2a,b + p,c + q by simplifying and solving the three equations.