IPMAT Indore 2025 (SA) - If the sum of the first 21 terms of the sequence: a/b, ab sqrt(b), ab^2, ab^2 sqrt(b), is a^mb^n, then the value of m+n is | PYQs + Solutions

IPMAT Indore 2025

Algebra

Progression & Series

Medium

If the sum of the first $21$ terms of the sequence: $\ln \frac{a}{b}, \ln \frac{a}{b \sqrt{b}}, \ln \frac{a}{b^{2}}, \ln \frac{a}{b^{2} \sqrt{b}}, \ldots$ is $\ln \frac{a^{m}}{b^{n}}$ , then the value of $m+n$ is $\qquad$

Entered answer:

✅ Correct Answer: 147

\ln \dfrac{a}{b}, \quad \ln \dfrac{a}{b\sqrt{b}}, \quad \ln \dfrac{a}{b^{2}}, \quad \ln \dfrac{a}{b^{2}\sqrt{b}}, \dots

The power of

b

in the denominator follows the pattern:

1, 1.5, 2, 2.5, \dots

This means the exponent increases by

0.5

each time.

For the

n

th term, the exponent of

b

in the denominator would be:

1 + (n-1) \times 0.5 = 0.5n + 0.5

So the

n

th term is:

\ln \dfrac{a}{b^{0.5n+0.5}} = \ln \dfrac{a}{b^{0.5(n+1)}}

To find the sum of the first 21 terms, we have:

\sum_{i=1}^{21} \ln \dfrac{a}{b^{0.5(i+1)}}

Using logarithm properties:

\ln \dfrac{a}{b^{0.5(i+1)}} = \ln a - \ln b^{0.5(i+1)} = \ln a - 0.5(i+1)\ln b

Now, we can calculate the sum:

\sum_{i=1}^{21} \ln \dfrac{a}{b^{0.5(i+1)}} = \sum_{i=1}^{21} (\ln a - 0.5(i+1)\ln b)

= 21\ln a - 0.5\ln b \sum_{i=1}^{21} (i+1)

= 21\ln a - 0.5\ln b \sum_{i=2}^{22} i

The sum of integers from 2 to 22 is:

\dfrac{(2+22)(22-2+1)}{2} = \dfrac{24 \cdot 21}{2} = 252

Therefore:

\sum_{i=1}^{21} \ln \dfrac{a}{b^{0.5(i+1)}} = 21\ln a - 0.5\ln b \cdot 252

= 21\ln a - 126\ln b

= \ln a^{21} - \ln b^{126}

= \ln \dfrac{a^{21}}{b^{126}}

\newline

Comparing with

\ln \dfrac{a^m}{b^n}

, we get

m = 21

and

n = 126

Therefore,

m + n = 21 + 126 = 147