JEE Main 2020 · 5 September · evening · Single correct · +4 / −1

If the sum of the first 20 terms of the series
log _(( 7^(1/2) ))x + log _(( 7^(1/3) ))x + log _(( 7^(1/4) ))x + ... is 460,
then x is equal to :

  1. (a) e²
  2. (b) 7 1/2
  3. (c) 7²
  4. (d) 7 46/21
Answer and solution

Answer (c) 7²

Solution

Maths sir · 4 steps

Pull the fractional power out of each base and every term becomes a whole-number multiple of log₇ x.

  1. Rewrite each term. Use log_(aᵏ) x = 1/k logₐ x. With k = 1/n this gives log_(7^(1/n)) x = n log₇ x.
  2. Find the multipliers. The first base is 7^(1/2), so n starts at 2. The 20th term has n = 21. The multipliers run 2, 3, 4, …, 21, an AP with common difference 1.

    (2 + 3 + … + 21)log₇ x = 20/2(2 + 21)log₇ x = 230 log₇ x

  3. Use the given sum. 230 log₇ x = 460, so log₇ x = 2.
  4. So. x = 7² = 49, which is option (c).

The idea to keep. A fractional power in the base of a log comes out as its reciprocal in front. Then always check where n starts before you add up the AP.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Sequences & Series

Asked about 250 times in JEE Main since 2020. This question is from 2020.