Question

If the p.th, q.th and r.th term of an A.P. as well as a G.P. (Same first term) are a, b and c respectively, prove that $a^{b-c}$ . $b^{c-a}$. $c^{a-b}$ =1

Answer 1

Answer:-

Given:

pth , qth , rth terms of an AP & GP are a , b, c.

By using common difference in AP,

→ c - b = b - a

Multiplying ( - 1) both sides,

→ b - c = a - b. -- equation (1)

→ 2b = a + c -- equation (2)

Again,

a , b , c are in GP,

Using. Geometric mean ,

→ b² = ac -- Equation (3)

Now,

We have to prove,

$\sf{ {a}^{b - c} \times {b}^{c - a} \times {c}^{a - b} = 1}$

Now Substitute eq- (1) here,

$→\sf{ {a}^{a - b} \times {c}^{a - b} \times {b}^{c - a} = 1}$

Using (aⁿ)*(bⁿ) = (ab)ⁿ,

$→ \sf{ {ac}^{a - b} \times {b}^{c - a} = 1 }$

Now ,

Substituting eq- (3) we get,

$→ \sf{ {{b}^{2}}^{(a - b)} \times {b}^{c - a} = 1}$

Now,

$\sf{ Using \:{{a}^{m} }^{n} = a^{mn}\: and\:a^{(m)} * a^{(n)} = a^{(m + n)},}$

$→\sf{ {b}^{2a - 2b + c - a} = 1} \\ \\ → \sf{ {b}^{a + c - 2b} = 1}$

Substitute eq - (2) here.

$→ \sf{ {b}^{2b - 2b} = 1} \\ \\→ \sf{ {b}^{0} = 1} \\ \\ →\sf{1 = 1}$

(a^0 = 1)

Hence, Proved.