Home › JEE Main › Mathematics › Sequences and Series › The 10th term of the AP $3, 7, 11, 15, \ldots$ is:
The 10th term of the AP $3, 7, 11, 15, \ldots$ is:
A$40$, mistakenly multiplying 10 by 4 (the difference)
B$39$, since $a_{10} = 3 + 9\cdot 4 = 39$ by $a + (n-1)d$
C$30$, taking 10 times the first term as the answer
D$43$, summing 3 and $10\cdot 4$ without the $-1$
Answer & Solution
Correct answer: B. $39$, since $a_{10} = 3 + 9\cdot 4 = 39$ by $a + (n-1)d$
$a_{10} = a + 9d = 3 + 36 = 39$.
Related questions
Special series cover natural numbers, their squares and their:If two positive numbers have A.M. 10 and G.M. 8, their sum is:The geometric mean of 4 and 16 is:The arithmetic mean of 4 and 16 is:The next term of the geometric progression 5, 10, 20 is:The common ratio of the progression 81, 27, 9, 3 is:The common ratio of the progression 3, 6, 12, 24 is:The numbers 2, 4 and 8 are consecutive terms of a: