Question

If a, b, c are in A.P. and ${{a}^{2}}$, ${{b}^{2}}$ and ${{c}^{2}}$ are in G.P. such that a < b < c and $a+b+c=\dfrac{3}{4}$ , then a is equals to
( a ) $\dfrac{1}{4}-\dfrac{1}{3\sqrt{2}}$
( b ) $\dfrac{1}{4}-\dfrac{1}{4\sqrt{2}}$
( c ) $\dfrac{1}{4}-\dfrac{1}{\sqrt{2}}$
( d ) $\dfrac{1}{4}-\dfrac{1}{2\sqrt{2}}$

Answer 1

Hint: To solve this question, we will first find the value of b and values of a and c in terms of value of b and common difference d. then, using concept of G.P, we will find the values of b and then by using value of b we obtained earlier, we will obtain values of d and then according to values of d, we will find the value of a.

Complete step-by-step answer:
Before we solve this question let us see what is an A.P and G.P.
A.P. is a special series of form a, a + d, a + 2d,….., a + ( n -1 )d where a is the first term of an A.P and d is a common difference between two consecutive terms.
${{n}^{th}}$ term of an A.P. is determined by ${{a}_{n}}=a+(n-1)d$ and summation of n terms of an A.P is $\dfrac{n}{2}(2a+(n-1)d)$ and also if a, b, c are in A.P and d is common difference of an A.P then,
 b - a = c – b = d.
G.P. is a special series of form $a,ar,a{{r}^{2}},a{{r}^{3}},........,a{{r}^{n-1}}$ where a is first term of G.P and r is common ratio between two consecutive terms.
${{n}^{th}}$ term of G.P. is determined by ${{a}_{n}}=a{{r}^{n-1}}$ and summation of n terms of G.P is $\dfrac{a(1-{{r}^{n}})}{(1-r)}$ and also if a, b, c are in G.P and r is common ratio of an G.P then, $\dfrac{b}{a}=\dfrac{c}{b}=r$ .
Now, in question it is given that a, b and c are in A.P and ${{a}^{2}}$, ${{b}^{2}}$ and ${{c}^{2}}$ are in G.P
So, we have b –a = c –b
On, re – arranging, we get
2b = c + a ,
 Now, we have $a+b+c=\dfrac{3}{4}$
Putting, value of a + c = 2b in $a+b+c=\dfrac{3}{4}$, we get
$b+2b=\dfrac{3}{4}$
$3b=\dfrac{3}{4}$
$b=\dfrac{3}{4\cdot 3}=\dfrac{1}{4}$
Now, as a, b and c are in A.P let common differences be d.
So, we can write a, b and c as b – d , b and b + d,
So, $a=\dfrac{1}{4}-d$ and $c=\dfrac{1}{4}+d$
Now, as ${{a}^{2}}$, ${{b}^{2}}$ and ${{c}^{2}}$ are in G.P
so, ${{({{b}^{2}})}^{2}}={{a}^{2}}{{c}^{2}}$
taking square root on both side we get
$\pm {{b}^{2}}=ac$
Or, $\pm {{b}^{2}}=\left( \dfrac{1}{4}-d \right)\left( \dfrac{1}{4}+d \right)$
Or, $\pm {{b}^{2}}=\left( \dfrac{1}{{{4}^{2}}}-{{d}^{2}} \right)$, using identity ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$
Now, as $b=\dfrac{1}{4}$
So, putting value of $b=\dfrac{1}{4}$, we get
$\dfrac{1}{{{4}^{2}}}=\left( \dfrac{1}{{{4}^{2}}}-{{d}^{2}} \right)$
\[{{d}^{2}}=0\]
\[d=0\], which means a = b = c but it is given that a < b < c, so it is not possible.
Now another case will be,
$-\dfrac{1}{{{4}^{2}}}=\left( \dfrac{1}{{{4}^{2}}}-{{d}^{2}} \right)$
\[{{d}^{2}}=\dfrac{1}{8}\]
\[d=\pm \dfrac{1}{2\sqrt{2}}\], but as a < b < c , so d > 0
So, \[d=\dfrac{1}{2\sqrt{2}}\]
Then, $a=\dfrac{1}{4}-\dfrac{1}{2\sqrt{2}}$ , as a = b - d

So, the correct answer is “Option d”.

Note: To solve this question one must know what is A.P and G.P and how the relation between the consecutive numbers. Always remember that, ${{n}^{th}}$ term of an A.P. is determined by ${{a}_{n}}=a+(n-1)d$ and summation of n terms of an A.P is $\dfrac{n}{2}(2a+(n-1)d)$and ${{n}^{th}}$ term of G.P. is determined by ${{a}_{n}}=a{{r}^{n-1}}$ and summation of n terms of G.P is $\dfrac{a(1-{{r}^{n}})}{(1-r)}$ . Also, as there is no direct value of any variable isn’t given so, while calculation consider each and every case you get while solving the equation.