# If the (a+1)th ,7th ,and (b+1)th terms of an A.P, are in G.P with a,6,b being in H.P, then the 4th term of this A.P is ? (A) -7/2 (B) 7/2 (C) 0 (D) 3

Apr 29, 2018

(C) 0.

#### Explanation:

For the AP in Question, suppose the ${n}^{t h}$ term and

the common difference are , resp., ${t}_{n} \mathmr{and} d \ne 0$.

If ${t}_{1} = x , \text{ then, } {t}_{n} = x + \left(n - 1\right) d , n \in \mathbb{N}$.

Given that, ${t}_{a + 1} , {t}_{7} , {t}_{b + 1}$ are in GP.

$\therefore \left(x + a d\right) , \left(x + 6 d\right) , \left(x + b d\right)$ are in GP.

$\therefore {\left(x + 6 d\right)}^{2} = \left(x + a d\right) \left(x + b d\right)$.

$\therefore \cancel{{x}^{2}} + 12 x d + 36 {d}^{2} = \cancel{{x}^{2}} + \left(a + b\right) x d + a b {d}^{2}$.

Dividing by $d \ne 0 , 12 x + 36 d = \left(a + b\right) x + a b d \ldots \ldots \left(\star\right)$.

Here, $a , 6 , b$ are in HP, or, $\frac{1}{a} , \frac{1}{6} , \frac{1}{b}$ in B

$\therefore 2 \cdot \left(\frac{1}{6}\right) = \frac{1}{a} + \frac{1}{b} , i . e . , \frac{1}{3} = \frac{a + b}{a b}$.

$\therefore a + b = \frac{1}{3} a b$.

Sub.ing in $\left(\star\right) , 12 x + 36 d = \frac{1}{3} a b x + a b d = \frac{1}{3} a b \left(x + 3 d\right)$

$\therefore 12 \left(x + 3 d\right) - \frac{1}{3} a b \left(x + 3 d\right) = 0$

$\therefore \left(x + 3 d\right) \left(12 - \frac{1}{3} a b\right) = 0$.

$\therefore \left(x + 3 d\right) = 0 , \mathmr{and} , a b = 36$.

In case, $a b = 36 , \text{ then } a + b = \frac{1}{3} a b = 12$.

Solving these for $a \mathmr{and} b$, we get, $a = b = 6$.

But, in that event, ${t}_{a + 1} = {t}_{b + 1} = {t}_{7}$ will make $d = 0$.

So, let us discard that possibility.

We are now left with $x + 3 d = 0$, which is, in fact, ${t}_{4}$.

Hence, the Right Option is (C) 0.

Enjoy Maths!