Logo
Jeeify
MCQEasyJEE 2024Arithmetic Progression (AP)

JEE Mathematics 2024 Question with Solution

The 20th20^{\text{th}} term from the end of the progression 20,19.25,18.5,17.75,…,−129.2520, 19.25, 18.5, 17.75, \ldots, -129.25 is:

  • A

    −118-118

  • B

    −110-110

  • C

    −115-115

  • D

    −100-100

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The progression is 20,19.25,18.5,17.75,…,−129.2520, 19.25, 18.5, 17.75, \ldots, -129.25.

Find: The 20th20^{\text{th}} term from the end.

First identify the common difference:

d=19.25−20=−0.75=−34d = 19.25 - 20 = -0.75 = -\frac{3}{4}

So this is an arithmetic progression with first term a=20a = 20 and common difference d=−34d = -\frac{3}{4}.

Now find the total number of terms using the last term −129.25=−5174-129.25 = -\frac{517}{4}:

an=a+(n−1)da_n = a + (n-1)d −5174=20+(n−1)(−34)-\frac{517}{4} = 20 + (n-1)\left(-\frac{3}{4}\right) −5174=804−34(n−1)-\frac{517}{4} = \frac{80}{4} - \frac{3}{4}(n-1) −5974=−34(n−1)-\frac{597}{4} = -\frac{3}{4}(n-1) n−1=199n-1 = 199 n=200n = 200

Reverse Progression Method

Treat the progression from the end as a new A.P. Then the first term of the reversed progression is −129.25=−5174-129.25 = -\frac{517}{4} and its common difference is 34\frac{3}{4}.

The 20th20^{\text{th}} term from the end is:

a20=−5174+(20−1)⋅34a_{20} = -\frac{517}{4} + (20-1)\cdot \frac{3}{4} a20=−5174+574a_{20} = -\frac{517}{4} + \frac{57}{4} a20=−4604a_{20} = -\frac{460}{4} a20=−115a_{20} = -115

Therefore, the 20th20^{\text{th}} term from the end is −115-115. The correct option is C.

The solution states option B, but its own working gives −115-115, which matches option C.

Common mistakes

  • Using the common difference as +34+\frac{3}{4} in the original progression is incorrect because the terms are decreasing. First compute the difference carefully from consecutive terms; here it is −34-\frac{3}{4}.

  • Taking the 20th20^{\text{th}} term from the end as the 20th20^{\text{th}} term from the beginning is wrong. If the total number of terms is nn, then the kthk^{\text{th}} term from the end is the (n−k+1)th(n-k+1)^{\text{th}} term from the start.

  • Converting −129.25-129.25 incorrectly to a fraction can spoil the calculation. Write it as −12914=−5174-129\frac{1}{4} = -\frac{517}{4} before substituting into the A.P. formula.

Practice more Arithmetic Progression (AP) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Arithmetic Progression (AP)

Related questions