级数 - 例题解析

答案 - B

解释

In the given A.P. we have a=5, d= (8-5) = 3 
∴ Tn= a+ (n-1) d= 5+ (n-1)3 = 3n+2 
T16= (3*16+2) = 50

答案 - A

解释

We have a =4 and d= (9-4) = 5 
Let the nth term 109. At that point 
(a+ (n-1) d= 109 ⇒ 4+ (n-1)*5 =109 
(n-1)*5= 105 ⇒ (n-1) = 21 ⇒ n= 22 
∴ 22nd term is 109.

答案 - A

解释

Let the given A.P contain A.P. contain n terms. At that point, 
A=7, d = (13-7)= 6 and Tn = 205 
∴ a+ (n-1) d =205 ⇒ 7+ (n-1)*6 = 198 ⇒ (n-1) =33 ⇒ n = 34 
Given A.P contains 34 terms. 

答案 - B

解释

Let, first term = a and normal contrast =d. 
T6 = 12 ⇒ a+5d= 12 …. (i) 
T8= 22 ⇒ a+7d = 22 … (ii) 
On subtracting (i) from (ii), we get 2d = 10 ⇒ d = 5 
Putting d= 5 in (i), we get a+5*5 = 12 ⇒ a= (12-25) =-13 
∴ First term = - 13, normal distinction = 5. 
T16= a+ 15d = - 13+15*5 = (75-13) = 62

答案 - C

解释

Here a =5, d= (9-5) = 4 and n = 17 
Sn = n/2[2a+ (n-1) d] 
S17 = 17/2 [2*5+ (17-1)*4] = (17/2*74) = 629

答案 - A

解释

Here a = 2, d = (5-2) = 3 and Tn = 182. 
Tn = 182 ⇒ a+ (n-1) d = 182 ⇒ 2+ (n-1)*3 = 182 ⇒ 3n = 183 ⇒ n= 61. 
Sn = n/2[2a+ (n-1) d] 
=61/2 {2*2+(61-1)*3} = (61/2* 184) = (61*92) = 5612. 

答案 - D

解释

Let the numbers be (a-d), an and (a+d). At that point, 
(a-d) +a+ (a+d) = 15 ⇒ 3a = 15 ⇒ a = 5 
(a-d)*a*(a+d) = 80 ⇒ (5-d)*5 * (5+d) = 80 
⇒ (25-d2) = 16 = d2 =9 ⇒ d = 3 
Numbers are 2, 5, 8 or 8, 5, 2. 

答案 - D

解释

Given numbers are in G.P in which a= 3 and r =6/3 = 2. 
∴ Tn = arn-1 ⇒ T9= 3*28 = (3*256) = 768 
Tn = 3*2n-1 = 6n-1

答案 - B

解释

Let A be the first term and r be the basic proportion. At that point, 
T4 = 54 ⇒ ar³ =54 ... (i) 
T4 = 13122 ⇒ ar8 = 13122 ...(ii) 
On isolating (ii) by (i) , we get r5 = 13122/54 = 243 =(3)5 ⇒ r =3 
Putting r =3 in (i), we get a*27 =54 ⇒ a = 2 
∴ First term =2 and common ratio =3.
T6= ar5 = 2*35= 486. Hence, 6th term = 486.