Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

51.

The sum of first n terms of an A.P. is `5n^2+3n` If its mth term is 168, (i) Find the value of m (ii) Find the 20th term of this AP

Answer» `s_n = 5n^2 + 3n`
`s_1 = 5(1)^2 + 3 xx 1 = 8`
`s_2 = 5(2)^2 + 3 xx 2 =20 + 6 = 26`
`s_3 = 5(3)^2 + 3 xx 3 = 45 + 9 = 54`
`8, 18,28, ....`
`a= 8`
`d=18-8=10`
`T_m = a +(m-1)d= 168`
`8 + (m-1) 10 = 168`
`m-1= 160/10`
`m=17`
`T_20 = a + (20-1)d = 8 + 19 xx 10`
`8+190= 198`
Answer
Discussion
52.

The ratio of the 11th term to the 18th term of an AP is `2:3`. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

Answer» Let a and d be the first term and common difference of an AP.
Given that, `a_(11) :a_(18)=2: 3`
`implies(a+10d)/(a+17d)=(2)/(3)`
`implies 3a +30d=2a+34d`
`impliesa =4d " "` ...(i)
Now, `a_(5)= a+4d=4d+4d=8d " "` [from Eq. (i)]
and`a_(21)= a+20d=4d+20d=24d " "` [from Eq. (i)]
` :. a_(5):a_(21)=8d:24d=1:3`
Now sum of the first five terms,
`S_(5)=(5)/(2)[2a+(5-1)d]`
` " " =(5)/(2)[2(4d)+4d] " "`[from Eq. (i)]
` " "=(5)/(2)[8d+4d]=(5)/(2)xx12d=30d`
and sum of the first 21 terms,
`S_(21)=(21)/(2)[2a+(21-1)d]`
` " " =(21)/(2)[2(4d)+20d] " "`[from Eq. (i)]
` " "=(21)/(2)(28d)=294d`
So, ratio of the sum of the first five terms to the sum of the first 21 terms
`S_(5):S_(21)=30d:294d=5:49`
Discussion
53.

The 4th term of an A.P. is three times the first and the 7thterm exceeds twice the third term by four. Find the first term and the commondifference.

Answer» Let the A.P be -`a,a+d,a+2d............`
according to the question,`a+3d=3a`...(1) and `a+6d=2(a+2d)+4`....(2)
From (1) and(2) `d=8 `and `a=12`
the A.P is `12,20,28,36,44........`
Discussion
54.

The 4th term of an A.P is equal to the 3 times ofthe first term and 7th term exceeds twice the 3rd term by 1. Find first term and common difference

Answer» Let the A.P be `a,a+d,a+2d.......` where `a` is the first term and d is the common difference.
Given`a_4=3a_1`=>`a+3d=3a`=>`3d=2a`....(1)
also `a_7=2a_3+1`=>`2d=a+1`...(2)
solving (1) and(2) we get `d=2` and `a=3`
Discussion
55.

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.(iii) The cost of digging a well after every metre of digging, when it costs Rs.150 for the first metre and rises by Rs. 50 for each subsequent metre.(iv) The amount of money in the account every year, when ` 10000 is deposited at compound interest at 8 % per annum.

Answer» `a_1=15`
`a_2=8`
`a_3=8`
`a_2-a_1=8-15=-7`
`a_3=a_2=8-8=0`
this is not a AP.
Discussion
56.

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer» Here, first term of AP,` a = 5`
Last termof AP,`l = 45`
Sum of AP, `S = 400`
We know,
`S = n/2(a+l) =>400 = n/2(45+5)=>n = 8**2 = 16`
So, number of terms is `16`.
As, last term is `45`, we can write,
`5+(16-1)d = 45`
`d = 40/15 = 8/3`
So, common difference is `8/3`.
Discussion
57.

The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.

Answer» Let the first term, common difference and the number of terms of an AP are a, d and n respectively.
Given that, first term (a) = - 5 and last term `(l)` = 45
Sum of the terms of the AP `= 120impliesS_(n)=120`
We know that, if last term of an AP is known, jthen sum of `n` terms of an AP is,
` " " S_(n)=(n)/(2)(a+l)`
`implies 120 =(n)/(2)(-5+45)implies12xx2=40xxn`
`implies n=3xx2impliesn=6`
` :.` Number of terms of an AP is known, then teh `n`th term of an AP is,
` l =a+(n-1)dimplies45= -5+(6-1)d`
`implies 50=5dimpliesd=10`
So, the common difference is 10.
Hence, number of terms and the common difference of an AP are 6 and 10 respectively.
Discussion
58.

Prove that the sum of `n`arithmetic means between two numbers in `n`times the single A.M. between them.

Answer» Let `A_1,A_2,A_3,..,A_n`
Am between a,b
`AP->a,A_1,A_2,...,A_n,b`
`b=a+(n+2-1)d`
`d=(b-a)/(n+1)-(A)`
Now,
`A_1+A_2+...+A_n=n/2[A_1+A_n]`
`a+b=A_1+A_n`
`A_1+A_22+...+A_n=n/2[a+b]`
`=n((a+b)/2)`
=n(AM between a and b).
Discussion
59.

Find an A.P. in which the sum of any number of terms is always threetimes the squared number of these terms.

Answer» AP-.a,d
`n/2[2a+(n-1)d]=3n^2`
`2a+(n-1)d=6n`
`2a+(d-6)n-d=0`
`d-6=0`
`d=6`
`2a-d=0`
`2a=6`
`a=3`.
Discussion
60.

If `(b-c)^2,(c-a)^2,(a-b)^2`are in A.P., then prove that `1/(b-c),1/(c-a),1/(a-b)`are also in A.P.

Answer» `2(c-a)^2=(b-c)^2+(a-b)^2`
`2/(c-a)=1/(b-c)+1/(a-b)`
`2/(c-a)=(b-c+a-b)/((b-c)(a-b)`
`2[(1-c)(a-b)]=-(a-c)^2`
`2[-b-ac+ab+bc]=-a^2-c^2+2ac`
`a^2+c^2-4ac=b^2-2ab+b^2-2bc`
`2[a^2+c^2]-4ac=b^2-2ab+a^2+b^2-2bc+c^2`
`2(a-c)^2=(b-a)^2+(b-c)^2`.
Discussion
61.

A,B and C completes a piece of work costing Rs. 1800. A worked for 6 days, B for 4 days and C for 9 days. If their daily wages are in the ratio 5:6:4, how much amount will be recieved by A.

Answer» Let daily wages of `A` is `5x`.
Then daily wages of `B` will be `6x` and daily wages for `C` will be `4x`.
Now, `A` worked for `6` days, `B` worked for `4` days and `C` worked for `9` days.
`:.` Total wages `= 6(5x)+4(6x)+9(4x) = 90x`
`:. 90x = 1800 => x = 20`
So, amount received by `A = 6(5**20) = 600` Rs.
Discussion