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Black Advanced Problems in Mathematics

Exam: Black Advanced Problems in Mathematics
Subject: Mathematics
Chapter: Sequences and Series

Black Advanced Problems in Mathematics · Mathematics · Sequences and Series

If (m+1)^ th , (n+1)^ th , and (r+1)^ th terms of a non-constant...

If

(m+1)^{\text{th}}

,

(n+1)^{\text{th}}

, and

(r+1)^{\text{th}}

terms of a non-constant A.P. are in G.P. and

m, n, r

are in H.P., then the ratio of first term of the A.P to its common difference is:

Options

A $-\frac{n}{2}$
B $-n$
C $-2n$
D $+n$

Correct answer

B. $-n$

Step-by-step solution

Let the A.P. be

A, A+D, A+2D, \dots

The terms are

T_{m+1} = A + mD

,

T_{n+1} = A + nD

,

T_{r+1} = A + rD

. Given they are in G.P.,

(A + nD)^2 = (A + mD)(A + rD)

.

A^2 + 2AnD + n^2D^2 = A^2 + A(m+r)D + mrD^22AnD - A(m+r)D = mrD^2 - n^2D^2A(2n - m - r)D = (mr - n^2)D^2

Since the A.P. is non-constant,

D \ne 0

.

\frac{A}{D} = \frac{mr - n^2}{2n - m - r}

Since

m, n, r

are in H.P.,

n = \frac{2mr}{m+r} \implies m+r = \frac{2mr}{n}

. Substitute

m+r

into the equation:

\frac{A}{D} = \frac{mr - n^2}{2n - \frac{2mr}{n}} = \frac{n(mr - n^2)}{2n^2 - 2mr} = \frac{n(mr - n^2)}{-2(mr - n^2)} = -\frac{n}{2}

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