In the infinite sequence 13, 17, 21, 25, 29, . . . , where each term

Question

In the infinite sequence 13, 17, 21, 25, 29, . . . , where each term is 4 greater than the previous term, the 46th term is

A. 183
B. 187
C. 191
D. 193
E. 197

CareerGeek · Answer

AP series
—> 46th term = a + (46 - 1)*d
= 13 + 45*4 = 193

Option D

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akdebonair · Answer

For AP series nth term is given as below.
T(n) = a + (n-1) d
T(n) = nth term
a = first term
d = difference between two terms

T(46) = 13 + 45*4= 193
IMO ans D

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Lampard42 · Answer

It is in an AP. So t46= a+45d=13+180=193 IMO D

EgmatQuantExpert · Answer

Solution 
  Given:  
 • An infinite sequence 13, 17, 21, 25, 29, . . . , where each term is 4 greater than the previous term

To find:  
 • The  46^{th}  term of the sequence

Approach and Working Out:   
 • First term, a = 13
• Common difference, d = 17 – 13 = 4 
• Therefore,  t_{46}  = a + (46 – 1) * d = 13 + 45 * 4 = 193

Hence, the correct answer is Option D.

Answer: D

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Archit3110 · Answer

an = a + (n-1)*d
a= 13 ; d= 4 , n = 46
so
we get a46 = 13+(45)*4 ; 193
IMO D

LeoN88 · Answer

Tn = a + (n-1)d
a= 13
n=46
d=4
Put values to get 193

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whyyash · Answer

Arithmetic progression formula => Tn = a + (n-1)d

=> n = 46
=> a = 13
=> d= 4

=> 13 + (46-1)*4 = 193

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