Explore topic-wise InterviewSolutions in Current Affairs.

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1.

We know that the sum of the interior angles of a triangle is`180^0`. Show that the sums of eth interior angles of polygons with `3,4,5,6,`sides for an arithmetic progression. Find the sum of the interior angles ofor a 21 sided polygon.

Answer» We know that sum of interior angles of a polygon of side `n=(2n-4)xx90^(@)=(n-2)xx180^(2)`
Sum of interior angles a polygon with sides 3 is 180.
Sum of interior angles of polygon with side `4=(4-2)xx180^(@)=360^(2)`
Similarly , sum of interior angles of polygon with side 5,6,7,. are `540^(@) ,720 ^(@),900^(@)`.......
the series will be `180^(@),360^(@),540^(@),720^(@),900^(@),......`
Here `a=180^(@)`
` and d=360^(2)-180^(@)=180^(@)`
since common difference is same between two consecutive terms of the series .
so ir from an AP .
we have to find the of interior angles of a 21 sides polygen .
it means , we have to find the 19 th term of the above series .
`a_(19)=a+(19-1)d`
`=180+18xx180=3420`
Discussion
2.

Any term of an AP (except first ) is equal to half the sum of terms which are equidistant from it .

Answer» True
Consider AnAP a,a,+d,A+2d ,…
Now `a_(2)+a_(4)=a+d+a+3d`
`=2a+4d=2a_(3)`
`a_(3)=(a_(2)+a_(4))/(2)`
Again ,` (a_(3)+a_(5))/(2)=(a+2d+a+4d)/(2)=(2a+6d)/(2)`
` =a+3d=a_(4)`
hence ,the statement is true .
Discussion
3.

If the sum of p terms of an AP is q and the sum of q terms is p, then show that the sum of `p+q` terms is `- (p+q)` , Also find the sum of first `p-q` terms (where , `pgtq`).

Answer» let first term common difference of the AP be and d respectively .
then `S_(p)=q`
`implies (P)/(2)[2a+(p-1)d]=q`
` 2a+(p-1)d=(2q)/(p)`
and `S_(q)-p`
`implies (q)/(2)[2a+(q-1)d]=p`
`implies 2a+(q-1)d=(2p)/(q)`
On subtracting Eq (ii) from Eq (i) we get
`2a+(p-1)d-2a-(q-1)d=(2q)/(p)-(2p)/(q)`
` implies [(p-1)-(q-1)]d=(2p^(2)-2p^(2))/(pq)`
`implies [p-1-q_1]d=(2(q^(2)-p^(2)))/(pq)`
`implies (p-q)d=(2(q^(2)-p^(2)))/(pq)`
`therefore d=(-2(p+q))/(pq)`
On substiuting the value of d in (eq (i) we get
` 2a+(p-1)((-2(p+q))/(pq))=(2q)/(p)`
`implies 2a=(2q)/(p)+(2(P+q)(P-1))/(pq)`
`implies a=[(2q)/(p) +(2(p+q)(p-1))/(pq)]`
`Now S_(p+q) -(p+q)/(2)[2a+(p+q-1)d]`
`=(p+q)/(2)[(2q)/(p)+(2(p+q)(p-1))/(pq)-((p+q-1)2(P+q))/(pq)]`
` =(p+q)[(q)/(p)+((p+q(p-1)-(p+q-1)(P+q))/(pq)]`
`=(p+q)[(q)/(p)+((p+q)(p-1)-(p+q-1)(P+q))/(pq)]`
`=(P+q)[(q)/(p)+((p+q)(p-1-p-q+1)(p+q))/pq)]`
`=p+q [(q)/(p)-(p+q)/(p)]=(p+q)[(q-p-q)/(p)]`
`S_(p+q)=-(p+q)`
`S_(p-q)=(p-q)/(2)[2a+(p-q-1)d]`
`=(p-q)/(2)[(2p)/(p)+(2(p+q)(p-1))/(pq)-(p-q-1)2(p+q))/(pq)]`
`=(p-q)[(q)/(p) +(p+q(p-1-p+q+1))/(pq)]`
`=(p-q)[(q)/(p)+((p+q)q)/(pq)]`
`=(p-q)[(q)/(p)+(p+q)/(p)]=(p-q)((p+2q))/(p)`
Discussion
4.

If the `p t h , q t h a n d r t h`terms of `a`G.P. are `a , b , c`respectively, prove that: `a^((q-r))dot^( )b^((r-p))dotc^((p-q))=1.`

Answer» Let A , D are the first term and common difference difference of Ap and x, R are the first term and common ratio fo Gp respectively .
According to the given condition ,
`A+(p-1)d=2`…(i)
`A+(q-1) d=b`***(iii)
` A +(r-1) d=c`...(iii)
and `a=xR^(p-1)` ...(iv)
`b=xR^(p-1)` ...(v)
`c=xR^(p-1)` ***(vi)
On subtracting Eq,(ii) from Eq (i) we get
`d(p-1-q+1)=a-b`
`a-b=d(p-q)`***(vii)
On subtracting Eq,(iii) from Eq (ii) we get
`d(p-1-r+1)=b-c`
`implies b-c=d(q-r)` ...(viii)
On subtracting Eq,(i) from Eq (iii) we get
`d(r-1-p+1)=c-a`
`c-a=d (r-p)` ....(ix)
taking LHS `=a^(b-c) b^(c-a) c^(a-b)`
using Eqs (iv) ,(v) ,(vi) and (viii) , (ix)
`LHS =(xR^(p-1))^(d(q-r) )(xR^(q-1) ) ^(d(r-p) ) (xR^(r-1))^(d(p-q))`
`=x^(d(q-1)+d(r-p)+d(p-q))R^((p-1)d(q-r)+d(r-p)+(r-1)d(p-q))`
`=x^(d(q-r+r=p-p-q))`
`R^(d(pq-pr-q+r+qr- pq -r +p rp -rp-p+q))=x^(0)r^(0)=1`
=RHS Hence proved .
Discussion
5.

The sum of first n terms of an AP is given by `S_n=2n^2 + 3n`. Find the comman difference of the AP.A. 3B. 2C. 6D. 4

Answer» Correct Answer - D
Given ` 1S_(n) = 3n+2n^(2)`
first term of the AP .
`T_(1) =3xx1+2(1)^(2)=3+2=5`
and `T_(1) ==s_(2)-S_(1)`
`=[3xx2+2xx(2)^(2)]-[3xx1+2xx(1)^(2) ]`
` =14-5=9`
`therefore ` Common difference `(d)=T_(2)-t_(1)=9-5=4`
Discussion
6.

If `1/a ,1/b ,1/c`are in A.P. then `(b+c)/a ,(c+a)/b ,(a+b)/c`are in :A. A.P.B. G.P.C. miscellaneousD. None of these

Answer» Correct Answer - A
N/a
Discussion
7.

If a, b, c are in A.P. as well as in G.P. then correct statement is :A. `a=b=c`B. `anebnec`C. `a=bnec`D. None of these

Answer» Correct Answer - A
N/a
Discussion
8.

If a, b, c are in A.P. and b-a, c-b, a are in G.P. then a:b:c=?A. `1:3:5`B. `1:2:3`C. `1:4:7`D. None of these

Answer» Correct Answer - B
N/a
Discussion
9.

If `x=1+a+a^(2)+...oo, a lt 1` and `y=1+b+b^(2)+...oo, b lt 1,` then `1+ab+a^(2)b^(2)+...oo :`A. `(x+y+xy)/(x+y-1)`B. `(x+y)/(x+y-1)`C. `(xy)/(x+y-1)`D. None of these

Answer» Correct Answer - C
N/a
Discussion
10.

let `S_(n)` denote the sum of the cubes of the first n natural numbers and `s_(n)` denote the sum of the first n natural numbers , then `sum_(r=1)^(n)(S_(r))/(s_(r))` equals toA. `(n(n+1)(n+2))/(6)`B. `(n(n+1))/(2)`C. `(n^(2)+3n+2)/(2)`D. None of these

Answer» Correct Answer - A
`sum _(r=1)^(n) (S_(r))/(S_(r))=(S_(1))/(S_(1))+(S_(2))/(S_(2))+(S_(3))/(S_(3))+...+(S_(n))/(S_(n))`
Let `T_(n)` be the nth term the above series .
`T_(n)=(S_(n))/(S_(n))=([(n(n+1))/(2)]^(2))/((n(n+1))/(2))`
`=(n(n+1))/(2)=(1)/(2)[n^(2)=n]`
`therefore `Sum of the above series `=sum T_(n)=(1)/(2)[sumn^(2)+sumn]`
`=(1)/(2)[(n(n+1)(2n+1))/(6)+(n(n+1))/(2)]=(1)/(2).(n(n+1))/(2)[((2n+1))/(3)+1]`
`=(1)/(4)m(n+1)[(2n+1+3)/(3)]=(1)/(4xx3)n(n+1)(2n+4)`
`=(1)/(12)n(n+1)(2n+4)=(1)/(6)n(n+1)(n+2)`
Discussion
11.

First negative term of the series `4,3(5)/(7),3(3)/(7),…` is :A. 15B. 16C. 17D. None of these

Answer» Correct Answer - B
N/a
Discussion
12.

The nth term of an A.P. is `(1)/(n)` and nth term is `(1)/(m).` Its (mn)th term is :A. mnB. `(1)/(mn)`C. 1D. None of these

Answer» Correct Answer - C
N/a
Discussion
13.

If `t_n`denotes the nth term of the series `2+3+6+11+18+...`then`t_(50)`=......`49^2-1`b. `49^2`c. `50^2+1`d. `49^2+2`A. `49^(2)-1`B. `49^(2)`C. `50^(2)+1`D. `49^(2)+2`

Answer» Correct Answer - D
Let `S_(n)` be sum of the series 2+3+6+11+18+***+`t_(50).`
`therefore s_(n) =2+3+6+11+18***+t_(50)`
`and S_(n) =0+2+3+6+11+18***+t_(49) +y_(50)`
On subtracting Eq (ii) from eq (i) we get
`0=2+1+3+5+7+***` upto 49 terms
`t_(50) =2+[1+3+55+7+***`upto 49 terms]
`=2+(49)/(2)[2xx1++48xx2]`
`=2+(49)/(2)xx[2+96]`
`=2+[49+49xx48]`
`2+49xx49=2+(49)^(2)`
Discussion
14.

The lengths of three unequal edges of a rectangular solids block are in GP . if the volume of the block is `216 cm^(3)` and the total surface area is `252cm ^(2)` then the length of the longest edge isA. 12cmB. 6 cmC. 18 cmD. 3 cm

Answer» Correct Answer - A
Let the length breadth and hight of rectangular solids block is `(a)/( r) ` a and ar respectively
`therefore "volume " =(a)/( r) xxaxxar =216 cm^(3)`
`implies a^(3) =216implies a^(3)=6^(3)`
`therefore a=6`
surface area ` =2((a^(2))/(r )+a^(2)r+a^(2))=252`
`implies 2a^(2)((1)/( r) _r+1)=252`
` implies 2xx36((1+r^(2)+r)/(r ))=252`
`implies (1+r^(2)+r)/(r ) =(252)/(2xx36)`
`implies 1+r^(2)+r=(126)/(36rimplies 1+r^(2)+r=(21)/(6)r`
`implies 6+6r^(2)=21rimplies 6r^(2)-15r+6=0`
`therefore r(1)/(2),2`
For ` r=(1)/(2) : "length"=(a)/(r)=(6xx2)/(1)=12`
Breadth `=a=6`
Height `=ar=6xx(1)/(2) =3`
for r=2 : length `=(a)/(r ) =(6)/(2)=3`
Breadth `=a=6`
height `=ar=6xx2=12`
Discussion
15.

if x , 2y and 3z are in AP where the distinct numbers x, yand z are in gp. Then the common ratio of the GP isA. 3B. `(1)/(3)`C. 2D. `(1)/(2)`

Answer» Correct Answer - B
Given x, 2y and 3z are in AP ,
then ,`2y=(x+3z)/(2)`
`implies y=(x+3z)/(4)`
`implies 4y=x+3z`
and x,y,z are in GP
then `(y)/(x)=(z)/(y) =lamda`
`implies y=x lamda and z=lamday - lamda^(2)x`
On subsitiuting these values in Eq (i) we get
`4(x, lamda )=x+3(lamda^(2)x)`
`implies 4lamdax=x+3lamda^(2)x`
`implies 4lamda=1+3lamda^(2)`
`implies 3lamda ^(2) -4lamda+1=0`
`=(3lamda-10(lamda-1)=0`
`therefore lamda =(1)/(3).lamda =1 `
Discussion
16.

The number of common terms in the series 17+21+25+…+417 and the series 16+21+26+…+466 are :A. 19B. 20C. 21D. None of these

Answer» Correct Answer - B
N/a
Discussion
17.

(ii)If the third term of G.P.is `4`, then find the product of first five termsA. `4^(3)`B. `4^(4)`C. `4^(5)`D. None of these

Answer» Correct Answer - C
It is given that `T_(3)=4`
let and r the term and common ratio respectively
then `ar^(2)=4`
Product of first 5 terms `=a.ar .ar^(2).ar^(3).ar^(4)`***(i)
`a^(5) r^(10)=(ar^(2))^(5)=(4)^(5)` [using Eq ,(i) ]
Discussion
18.

If the `p^(t h)a n dq^(t h)`terms of a G.P. are `qa n dp`respectively, show that `(p+q)^(t h)`term is `((q^p)/(p^q))^(1/(p-q))`.

Answer» Let the first term and common ratio of GP be and r, respectively .
Accodring to the question , pth term = q
`implies a.r^(p-1)=q`
and q th term =p
`implies ar^(q-1)=p`
On dividing Eq (i) by Eq (ii) We get
`(ar^(p-1))/(ar^(q-1))=(q)/(p)`
` implies r^(p-1-q+1)=(q)/(p)`
`implies r^(p-q)=(q)/(p)implies r((q)/(p))^((1)/(p-q))`
On substituting the value of r in Eq (i) , we get
`a((q)/(p))^((p-1)/(p-q))=qimplies a=(q)/(((q)/(p))^((p-1)/(p-q)))=q. ((p)/(q))^(^(p-1)/(p-q))`
`therefore (p+q) "th term "T_(p+q)= a.r ^(p+q-1)=q/((p)/(q))^((p-1)/(p-q)). (r) ^(p+q-1)`
`=q((p)/(q))^((p-1)/(p-q))[((q)/(p))^((1)/(p-q))]^(p+q-1)=q.((p)/(q))^((p-1)/(p-q))((q)/(p))^((p+q-1)/(p-q))`
` =q.((p)/(q))^((p-1)/(p-q))((p)/(q))^((-(p+q-1))/(p-q))=q.((p)/(q))^((p-1)/(p-q)-((p+q-1))/(p-q))`
`=q.((p)/(q))^((p-1-p-q+1)/(p-q))
=q.((p)/(q))^((p-1-p-q+1)/(p-q))=q./((p)/(q))^((-q)/(p-q))`
` a=q.((p)/(q))^((p-1)/(p-q))`
Now (p+q) th term i.e., `a_(p+q)=ar^(p+q-1)`
`=.((p)/(q))^((p-1)/(p-q)).((q)/p)^((p+q-1)/(p-q))`
` =q.q^((p=q-1-p+1)/(p-q))/(p^((p+q-1-p+1)/(p-q)))=q.(q^((q)/(p-q))/(p^((q)/(p-q))))`
Discussion
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