Following quiz provides Multiple Choice Questions (MCQs) related to Progression. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Which term of the A.P. 2, 7, 12, 17...are 87?
Here a =2, d= (7-2) =5. Let the nth term be 87. Then, a + (n-1) d =87 ⇒ 2+ (n-1)*5 = 87 ⇒ (n-1)*5= 85 ⇒ (n-1) = 17 ⇒ n = 18 ∴ 18th term is 87.
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:
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 - Aggregate number of whole numbers somewhere around 100 and 200 which are distinct by both 9 and 6 is:
LCM of 9 and 6 = 18 So, each one must be multiple of 18. Requisite no. is 108,126, 144, 162, 180, 198. Their no. is 6.
This is an A.P. in which a = 5, d =1 and L=19. Let the number of its term be n. Then, Tṇ = 19 ⇒ a + (n-1) d =19 ⇒ 5 + (n-1) * 1 = 19 ⇒ (n-1) =14 ⇒ n= 15. ∴ Sṇ = n/2 * (a+L) = 15/2 *(5+19) = 180.
Sum = (1+3 +5+7+...to 20 terms Here a = 1, d = (3-1) =2 and n = 20. ∴ Sṇ = n/2 {2a + (n-1) d} =20/2 *{2*1+ (20-1)*2} =10*40=400.
Let Tn = 1280. Then ar (ⁿ⁻ⁱ) = 1280 ⇒ Here a= 5 and r =2 ∴ 5*2(ⁿ⁻ⁱ) =1280 ⇒256 = 2(ⁿ⁻ⁱ) =2⁸ ⇒ n-1 =8 ⇒ n= 9
Q 7 - The second term of a geometrical progression is 2/3 and its fifth term is 16/81. Its seventh term is:
Let the G.P. be a, ar, ar2... Then T2 =2/3 and T₅ = 16/81 ⇒ ar = 2/3 ...(i) And ar⁴= 16/81 ...(ii) ∴ar⁴/ar =16/81*3/2 = 8/27 ⇒ r3 = 8/27 ⇒ r= 2/3 Putting r =2/3 in (i), we get =a= 1. T7 = ar⁶ = 1* (2/3)⁶ = 2⁶/3⁶ = 64/729
Q 8 - A man heaps logs of wood so that the top layer contains one log and every lower layer has one more than the layer above. In the event that there are 15 layers, the aggregate number of logs will be:
Total number of logs = 1+2+3+...+15. This is an A.P. in which a =1, d =1 and n= 15. Sn = n/2(a+1) = 15/2 (1+15) =120
Q 9 - The estimation of (13+23+33+? ? ..+153) - (1+2+3+...+15) =?
(13+23+33+...+153) - (1+2+3+...+15) = (15*16/2)2 -1/2 *15*16 = (120)2 -120 =(14400- 120) = 14280.