Question

Find the sum of first nth term of an AP series is ㏒a+㏒a²\b+㏒a³\b²+........

Answer 1

Answer:

log [ a^{ ( n² + n )/2 } ] / [ b^{ ( n² - n )/2 } ]

Step-by-step explanation:

log a + log a²/b + log a³/b² + ..............

It can be written as

log a¹/b⁰ + log a²/b¹ + log a³/b² + ....... upto nth term log aⁿ / bⁿ⁻¹  

From this we can understand that power of ' a ' and 'b' are increasing uniformly. So we can say that they are in AP

= log a + log a²/b + log a³/b² + ........ log aⁿ/bⁿ⁻¹

Using Product rule log a + log b = log ab

= log ( a × a² × a³ ........ aⁿ ) / ( b × b² × b³ ..... bⁿ⁻¹ )

Using yᵃ × yᵇ = yᵃ ⁺ ᵇ

= log { a^( 1 + 2 + 3 + ..... n ) } / { b^( 1 + 2 + 3 + ..... n-1 ) }

Now using Sum of n terms of AP formula for powers

= log [ a^{ n/2 × ( 1 + n ) } ] / [ b^{ ( n - 1 )/2 × ( 1 + n - 1 ) } ]

= log [ a^{ ( n² + n )/2 } ] / [ b^{ ( n² - n )/2 } ]

Hence we found the required answer.

Answer 2

ANSWER:

Given

AP = log a + log a²/b + log a³/b² + .........

♦ Using

• b⁰ = 1
• a = a¹
• b = b¹

We get

→ log a¹/b⁰ + log a²/b¹ + log a³/b²

Upto nth term

→ log a¹/b⁰ + log a²/b¹ + log a³/b² +....+ log aⁿ/bⁿ-¹

We can understand that a/b is increasing uniformly

Using

♦ log a + log b = log ab

→ log [(a¹/b⁰)(a²/b¹)(a³/b²) ......... (aⁿ/bⁿ-1)

→ log [(a¹ × a² × a³ ..... aⁿ)/(b⁰ × b¹ × b² × ...... × bⁿ-¹)

Using

aⁿ × a^m = a^(m + n)

→ log [(a¹+²+³+……ⁿ)/(b⁰+¹+²+……ⁿ-¹)

Using

♦ Sn = n/2[2a + (n - 1)d]

→ log [a^{ n/2 x (1+ n ) }]/[b^{ ( n - 1)/2 x (1+n-1)}]

→ log (a^{(n² + n )/2}]/[b^{(n² - n )/2]]