Q 1 - What number of term arrives in the A.P. 7, 11, 15..... 139?
A. 33
B. 34
C. 32
D. 31
Here a = 7 and d = (11-7) = 4 Let there be n terms in the given A.P. then, Tn = 139 ⇒a+ (n-1) d =139 ⇒ 7+ (n-1)*4 = 139 ⇒ (n-1) *4 = 132 ⇒ (n-1) = 33 ⇒ n = 34
Q 2 - In the event that the fourth term of a number juggling movement is 14 and its twelfth term is 70, then its first term is:
A. -10
B. -7
C. 7
D. 10
Let the first term of the A.P. be a and common difference be d. then, a+ 3d = 14 ...(i) a+11d = 70 ...(ii) On subtracting (i) from (ii), we get 8d =56 d = 7 Putting d = 7 in (i), we get a+3*7=14 ⇒ a= (14-21) = -7 ∴ First term = -7
Q 3 - What number of products of arrive between the whole numbers 15 and 105, both comprehensive?
A. 30
B. 31
D. 33
Requisite no. are 15, 18, 21, 24, 105. Here a = 15 and d = (18-15) = 3 Let their number be n. Then, a + (n-1) d =105 ⇒ 15+ (n-1)*3 = 105 ⇒ a+ (n-1) d = 105 ⇒ 15+ (n-1)*3 = 105 ⇒ (n-1)*3 = 90 ⇒ n-1 =30 ⇒ n = 31
Q 4 - The total of the all odd number somewhere around 100 and 200 is:
A. 3750
B. 6200
C. 6500
D. 7500
Requisite sum = 101 +103 + 105+...+199. This is an A.P. in which a = 101, d =2 and L= 199. A + (n-1) d =199 ⇒ 101 + (n-1) *2 = 199 ⇒ (n-1) * 2 = 98 ⇒ (n-1) = 49 ⇒ n = 50. ∴ Sum = n/2 * (a+L) = 50/2 * (101 +199) = (50*150) =7500.
Q 5 - The total of every single common number from 75 to 97 is:
A. 1598
B. 1798
C. 1958
D. 1978
Sum = 75+76+77+...+97. Here a =75, d = (76-75) =1 Let the number of terms be n. Then, A+ (n-1) d =97⇒ 75 + (n-1)*1 =97 ⇒ (n-1) = 22 ⇒ n= 23. ∴ Sum = 23/2 (75 + 97) = (23/2 *172) = (23 *86 ) = 1978.
Q 6 - In the event that the fourth and ninth term of a G.P. is 54 and 13122 separately, there its second term is:
A. 6
B. 12
C. 18
D. 9
Let its 1st term be a and common ratio r. Then, ar3 = 54 and ar⁸ = 13122 ∴ ar⁸/ ar3 = 13122/54 ⇒ r⁵ =243 = 3⁵ =r = 3 ∴ a* 33 = 54 ⇒ a*27 = 54 = > a =2 2nd term = ar = (2*3) =6
Q 7 - Three numbers are in G.P. Their whole is 28 and item is 512. The numbers are:
A. 2,6,18
B. 2,8,16
C. 4,8,16
D. 6,9,13
Let the number be a/r, a, ar. Then, a/r* a*ar = 512 ⇒ a3 =83 ⇒ a =8 8/r+8+8r =28 ⇒ 8/r +8r =20 ⇒ 2/r+2r =5 ⇒ 2r2-5r+2 =0 ⇒ 2r2-4r-r+2 =0 ⇒ 2r (r-2)- (r-2)=0 ⇒ (r-2)(2r-1) =0 ⇒r = 2 or r = 1/2 Numbers are 4,8, 16.
Q 8 - A few buys National Savings Certificates each year whose worth surpasses the earlier years buy by Rs 400. Following 8 years, she finds that she has obtained declarations whose aggregate face worth is Rs 48000. What is the face estimation of the Certificates acquired by her in the first years?
A. Rs 4300
B. Rs 4400
C. Rs 4500
D. Rs 4600
Let the required value be Rs a. Also d= 400, n = 8 and Sn = 48000. Sn = n/2 [2a + (n -1) d] ⇒ 8/2 *[2a + 7 *400] = 48000 ⇒ 2a + 2800 = 12000⇒ 2a=9200⇒ a = 4600
Q 9 - (142 +152 +......+302) =?
A. 3836
B. 8336
C. 8366
D. 8636
We know that (12+22+32+... +a2) = {n(n+1)(2n+1)}/6 Given Exp. = (12+22+...+132+142+??+302)-(12+22+...+132) = (30*31*61)/6 - (13*14*27)/6 = (9455- 819) = 8636
Q10 Find the position of 62 in the following series 2, 5, 8, ....?
A. 26
B. 23
C. 21
D. 20