220 + Interview Questions in ARITHMETIC-PROGRESSION IN GENERAL AWARENESS Page 5 InterviewSolution

201.	What is the sum of the first seven multiples of 5 ?A. 35B. 140C. 280D. 49
Answer» Correct Answer - B `{S_(n) = ( n )/(2) ( a+l) = ( 7)/(2) ( 5+ 35) }`

Discussion

202.	For an A.P. if `t_(8) =6, d = - 2 ` , then what is the value of a ?A. 18B. 14C. 16D. 20
Answer» Correct Answer - D `[t_(n) =a+ (n-1) d]`

Discussion

203.	If for an A.P., d=10 , what is the value of `t_(6)-t_(2)` ?A. 10B. 20C. 30D. 40
Answer» Correct Answer - D `[t_(x) -t_(y) = ( x-y) d ]`

Discussion

204.	Which of the following is the sum of the first 30 natural nambers?A. 464B. 465C. 462D. 461
Answer» Correct Answer - B `{S_(n) = ( n )/(2) [ 2a+ ( n-1) d] ` OR` S_(n) = ( n( n +1))/(2)}`

Discussion

205.	Find the sum of two middle terms of the AP `-4/3,-1,-2/3,-1/3,...,4(1/3)`
Answer» Here, `a=-(4)/(3), d=-1-(-(4)/(3))=(1)/(3)` Let,` a_(n)=4(1)/(3)` `rArr (-4)/(3)+(n-1)(1)/(3)=(13)/(3) rArr (n-1)(1)/(3)=(17)/(3)` `rArr n-1=17 rArr n=18` Now, the two middle most terms are `a_(9) and a_(10)`. `:. A_(9)+a_(10)=a+8d+a+9d` `=2a+17d` `=2(-(4)/(3))+17((1)/(3))=(-8)/(3)+(17)/(3)=3`

Discussion

206.	Jaspal Singh repays his total loan of Rs. 118000 by paying every month starting with the first installment of Rs. 1000. If he increases the installment by Rs. 100 every month, what amount will be paid by him in the 30th installment? What amount of loan does he still have to pay after the 30th installment?
Answer» Here, a=Rs. 1000, d=Rs. 100 `:. a_(30)=a+29d` `=Rs. (1000+29xx100)=Rs. 3,900` `rArr` Amount paid in 30th instalment = Rs. 3,900 and `S_(30)=Rs. (30)/(2)(2a+29)` `=Rs. 15(2000+29xx100)=Rs. 15xx 4900=Rs. 73,500` `rArr` Amount still have to pay after 30th instalment `=Rs. 1,18,000-Rs. 73,500=Rs. 45,500`

Discussion

207.	Find the value of x for which `(x+2),2x(2x+3)` are three consecutive terms of A.P.
Answer» Correct Answer - x =5

Discussion

208.	If the numbers a, 9, b, 25 form an AP, find a and b.
Answer» Correct Answer - a = 1, b = 17 We have, 9-a = b-9 = 25-b `therefore 2b = 34 rArr b = 17` Also, 9-a = 17 -9 = 8 `rArr` a = 9-8=1.

Discussion

209.	If 18, a, (b-3) are in AP, then find the value of (2a-b).
Answer» Correct Answer - 15 We have, a - 18 = (b-3-a) `rArr` 2a-b = (18-3) = 15

Discussion

210.	If `k , 2k-1`and `2k+1`are threeconsecutive terms of an A.P., the value of `k`is`-2`(b) `3`(c) `-3`(d) `6`
Answer» Correct Answer - k = 3 We have, (2k-1) -k = (2k +1) - (2k -1) `rArr k -1 = 2 rArr k = 3.`

Discussion

211.	The firstthree terms of an A.P. respectively are `3y-1, 3y+5`and `5y+1`. Then, `y`equals`-3`(b) `4`(c) `5`(d) `2`
Answer» Correct Answer - y = 5 We have, (3y+5) - (3y-1) = (5y +1) - (3y+5) `rArr 2y-4 = 6 rArr 2y = 10 rArr y = 5`

Discussion

212.	A contract on construction job specifies apenalty for delay of completion beyond a certain date as follows: Rs 200 forthe first day, Rs 250 for the second day Rs 300 for the third day, etc., thepenalty for each succeeding day being Rs 50 more
Answer» Correct Answer - ₹ 27750 Total penalty = ₹ (200 + 250 + 300 +… up to 30 terms). `T_(30) = (200 + 29 xx 50) = 1650` `therefore "required sum" = ₹ {(30)/(2) (200 + 1650)} = ₹27750`

Discussion

213.	If `a_1,a_2,a_3, ,a_n`are in A.P. withcommon difference `d(w h e r ed!=0)`, then the sum ofseries.`s in dd(cos e ca_1cos e ca_2+cos e ca_2cos e ca_3+=cos e ca_(n-1)cos e ca_n)`is equal to`cota_1-cota_ndot`
Answer» Since ` a_(1) , a_(2),a_(3) ,….,a_(n)` are in AP with common difference d, we have ` (a_(2)-a_(1)) = (a_(3)-a_(2)) = (a_(4)-a_(3))= ….= (a_(n) - a_(n-1)) =d ` Sin d ` ( cosec a_(1) coses a_(2) + cosec a_(2) cosec a_(3) +……+ cosec a_(n-1) cosec a_(n)` ) ` sind/(sina_(1)sina_(2))+ sind/(sina_(2) sina_(3))+sind/(sina_(3)sina_(4))+......+sindd/(sina_(n-1)sina_(n))` ` sin (a_(2) -a_(1))/(sina_(1) sin a_(2)) + sin (a_(3) -a_(2))/ (sin a_(2) sina_(3)) + sin( a_(4) -a_(3))/(sin a_(3) sin a_(4))/+ ....+ sin(a_(n)-a_(n-1))/(sin a_(n-1) sin a_(n)) ` ` [ therefore d= (a_(2) -a_(1) = ( a_(3) -a_(2)) =...= (a_(n) -a_(n -1)]` ` ( sina_(2) cos a_() - cos a_(2) sin a_(1))/(sin a_(1) sin a_(2)) + (sin a_(3) cos a_(2) -cos a_(3) sin a_(2))/ (sin a_(2) sin a_(3)) + ( sin a_(n) cos a_(n-1) -cos a_(n) sin a_(n-1))/ (sin a_(n-1) sina_(n))` ` = ((cos a_(1))/ (sin a_(1)) - (cos a_(2))/sin a_(2)) + ((cos a_(2))/(sin a_(2)) - (cos a_(3))/ (sin a_(3))) +......+ ((cos a_(n-1))/(sin a_(n-1)) - (cos a_(n))/(sin a_(n)))` ` ( cot a_(1) - cot a_(2)) + ( cot a_(2) -cot a_(3)) + (cot a_(3) - cot a_(4)) + .....+ ( cot a_(n-1) -cot a_(3)) + (cot a_(3) - cot a_(4)) + ....+ (cot a_(n-1) - cota_(n))` ` = ( cot a_(1) -cot a_(n))` Hence, sin d ` (coses a_(1) cosec a_(2) + cosec a_(2) cosec a_(3) + .....+ cosec a_(n-1) cosec a_(n)) = ( cot a_(1) -cot a_(n))`

Discussion

214.	If `a_1,a_2,a_3, ,a_n`are in A.P., where `a_i >0`for all `i`, show that`1/(sqrt(a_1)+sqrt(a_2))+1/(sqrt(a_1)+sqrt(a_3))++1/(sqrt(a_(n-1))+sqrt(a_n))=(n-1)/(sqrt(a_1)+sqrt(a_n))dot`
Answer» Let d be the common difference of the given AP. Then , ` ( a_(2) -a_(1)) = (a_(3) -a_(2)) = …..= (a_(n) -a_(n-1))= d ` Now , LHS ` 1/ ( (sqrta_(2) + sqrta_(1))) + 1 / (( sqrta_(3) + sqrta_(2)))+…..+ 1/((sqrta_(n) + sqrta_(n-1)))` ` ((sqrta_(2) - sqrta_(1)))/((sqrta_(2) +sqrta_(1))sqrta_(2) -sqrta_(1))) + ((sqrta_(3)-sqrta_(2)))/((sqrta_(3) + sqrta_(2)) (sqrta_(3) -sqrta_(3) -sqrta_(2))) +......+ ((sqrta_(n) -sqrta_(n-1)))/((sqrta_(n)+sqrt(a_(n-1))) (sqrta_(n) -sqrt(a_(n-1)))` `((sqrta_(2)-sqrta_(1)))/((a_(2)-a_(1)))+((sqrta_(3)-sqrta_(2)))/((a_(3)-a_(2)))+.......+((sqrta_(n)-sqrta_(n-1)))/((a_(n) -a_(n-1)))` ` ((sqrta_(2) -sqrta_(1))/d + ((sqrta_(3))-sqrta_(2))/d+....+ ((sqrta_(n) -sqrta_(n-1)))/d` `[ therefore (a_(2) -a_(1)) = (a_(3) -a_(2)) = (a_(4) -a_(3)) = .......= (a_(n) -a_(n-1)) =d]` ` 1/d .(sqrta_(n) -sqrt(a_(1)) + (sqrta_(3) -sqrt(a_(2)) + (sqrta_(4)-sqrta_(3)) +.....+ (sqrta_(n) -sqrta_(n-1))} ` =` 1/d .(sqrta_(n) -sqrt(a_(1))= 1/d .{(sqrta_(n) -sqrt(a_(1))xx ((sqrta_(n) +sqrta_(1)))/(sqrta_(n) + sqrt(a_(1)))}` ` 1/d . {((a_(n)-a_(1))/((sqrta_(1) +sqrta_(n))] = 1/d. {(a_(1) + (n+1)_ d-a_(1))/(sqrta_(1) + sqrta_(n))}` ` [ a_(n) =a_(1) + ( n-1) d]` `1/d .{ ((n-1) d)/( sqrta_(1) +sqrta_(n)))} = ((n-1))/((sqrta_(1)+sqrta_(n))) `= RHS Thus, LHS=RHS and hence the result follows.

Discussion

215.	The sum of first `n` terms of an AP is `(3n^2+6n)` Find the `n^(th)` term and `15^(th)` terms of the AP
Answer» Correct Answer - `T_(n) = (6n + 3), T_(15) = 93`

Discussion

216.	Find the sum of 16 terms of the AP , ` 6, 5""1/3,4""2/3,4`,…….
Answer» Correct Answer - 16

Discussion

217.	Find the sum of 23 terms of the AP 17,12,7,2,-3,….
Answer» Correct Answer - -874

Discussion

218.	If the sum of first 7 terms of an AP is 49 and that of 17 terms is289, find the sum of first n terms.
Answer» Correct Answer - `S_(n) = n^(2)`

Discussion

219.	Find the sum of all multiples of 7 lying between 500 and 900.
Answer» All multiples of 7 lying between 500 and 900 are 504, 511, 518, …, 896 This is an AP in which a = 504, d = 7 and l = 896. Let the given AP contain n terms. Then, `T_(n) = 896 rArr a+ (n-1)d = 896 rArr 504 + (n-1) xx 7 = 896` `rArr 497 + 7n = 896 rArr 7n = 399 rArr n = 57.` `therefore "required sum" = (n)/(2) (a+l)` ` = (57)/(2) *(504 + 896) = ((57)/(2) xx 1400)` = 39900.` Hence, the required sum is 39900.

Discussion

220.	Find the (i) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5. (ii) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5. (iii) sum of those integers from 1 to 500 which are multiples of 2 or 5.
Answer» Multiples of 2 from 1 to 500 are 2, 4, 6, ..., 500 (A.P.) Here, a=2, d=4-2=6-4=2 Let `a_(n)=500` `rArr 2+(n-1)2=500 rArr n=250` and `S_(250)=(250)/(2)(2+500)=125xx502=62750` Multiples of 5 from 1 to 500 are 5, 10, 15, ... (A.P) Here, `a=5, d=10-5=15-10=5` Let ` a_(n)=500` `rArr 5+(n-1)5=500 rArr n=100` and `S_(100)=(100)/(2)(5+500)=50xx505 = 25250` L.C.M. of 2 and 5 is 10 `:.` Multiples of 10 from 1 to 500 are 10, 20, 30, ... 500 (A.P) Here, `a=10, d=20-10=30-10=10` Let `a_(n)=500` `rArr 10(n-1)10=500 rArr n=50` and `S_(50)=(50)/(2)(10+500)=25xx510=12750` Now, the sum of all integers from 1 to 500 and multiples of 2 or 5 =sum of multiples of 2 + sum of multiples of 5 - sum of multiples of 10 ` = 62750+25250-12750=75250`

Discussion

Explore topic-wise InterviewSolutions in .

What is the sum of the first seven multiples of 5 ?A. 35B. 140C. 280D. 49

For an A.P. if `t_(8) =6, d = - 2 ` , then what is the value of a ?A. 18B. 14C. 16D. 20

If for an A.P., d=10 , what is the value of `t_(6)-t_(2)` ?A. 10B. 20C. 30D. 40

Which of the following is the sum of the first 30 natural nambers?A. 464B. 465C. 462D. 461

Find the sum of two middle terms of the AP `-4/3,-1,-2/3,-1/3,...,4(1/3)`

Find the value of x for which `(x+2),2x(2x+3)` are three consecutive terms of A.P.

If the numbers a, 9, b, 25 form an AP, find a and b.

If 18, a, (b-3) are in AP, then find the value of (2a-b).

If `k , 2k-1`and `2k+1`are threeconsecutive terms of an A.P., the value of `k`is`-2`(b) `3`(c) `-3`(d) `6`

The firstthree terms of an A.P. respectively are `3y-1, 3y+5`and `5y+1`. Then, `y`equals`-3`(b) `4`(c) `5`(d) `2`

A contract on construction job specifies apenalty for delay of completion beyond a certain date as follows: Rs 200 forthe first day, Rs 250 for the second day Rs 300 for the third day, etc., thepenalty for each succeeding day being Rs 50 more

If `a_1,a_2,a_3, ,a_n`are in A.P. withcommon difference `d(w h e r ed!=0)`, then the sum ofseries.`s in dd(cos e ca_1cos e ca_2+cos e ca_2cos e ca_3+=cos e ca_(n-1)cos e ca_n)`is equal to`cota_1-cota_ndot`

If `a_1,a_2,a_3, ,a_n`are in A.P., where `a_i >0`for all `i`, show that`1/(sqrt(a_1)+sqrt(a_2))+1/(sqrt(a_1)+sqrt(a_3))++1/(sqrt(a_(n-1))+sqrt(a_n))=(n-1)/(sqrt(a_1)+sqrt(a_n))dot`

The sum of first `n` terms of an AP is `(3n^2+6n)` Find the `n^(th)` term and `15^(th)` terms of the AP

Find the sum of 16 terms of the AP , ` 6, 5""1/3,4""2/3,4`,…….

Find the sum of 23 terms of the AP 17,12,7,2,-3,….

If the sum of first 7 terms of an AP is 49 and that of 17 terms is289, find the sum of first n terms.

Find the sum of all multiples of 7 lying between 500 and 900.

Find the (i) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5. (ii) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5. (iii) sum of those integers from 1 to 500 which are multiples of 2 or 5.