Quiz Discussion

If an A.P. has a = 1, tn = 20 and sn = 399, then value of n is :

Course Name: Quantitative Aptitude

  • 1] 20
  • 2] 32
  • 3] 38
  • 4] 40
Solution
No Solution Present Yet

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# Quiz
1
Discuss

In an A.P., if d = -4, n = 7, an = 4, then a is

  • 1]

    6

  • 2]

    7

  • 3]

    20

  • 4]

    28

Solution
2
Discuss

If a, b, c are in A.P., then (a – c)2/ (b2 – ac) =

  • 1]

    3

  • 2]

    4

  • 3]

    1

  • 4]

    2

Solution
3
Discuss

15th term of A.P., x - 7, x - 2, x + 3, ........ is

  • 1] x + 63
  • 2] x + 73
  • 3] x + 83
  • 4] x + 53
Solution
4
Discuss

What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

  • 1]

    897

  • 2]

    1,64,850

  • 3]

    1,64,749

  • 4]

    1,49,700

Solution
5
Discuss

If 18, a, b - 3 are in A.P. then a + b =

  • 1] 19
  • 2] 7
  • 3] 11
  • 4] 15
Solution
6
Discuss

If the sum of the series 2 + 5 + 8 + 11 … is 60100, then the number of terms are

  • 1]

    100

  • 2]

    150

  • 3]

    200

  • 4]

    250

Solution
7
Discuss

If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is

  • 1] 2
  • 2] 3
  • 3] 1
  • 4] 4
Solution
8
Discuss

If S1 is the sum of an arithmetic progression of ‘n’ odd number of terms and S2 is the sum of the terms of the series in odd places, then \(\frac{{{S_1}}}{{{S_2}}}\)

 

  • 1]

    \(\frac{{2n}}{{n + 1}}\)

  • 2]

    \(\frac{n}{{n + 1}}\)

  • 3]

    \(\frac{{n + 1}}{{2n}}\)

  • 4]

    \(\frac{{n - 1}}{n}\)

Solution
9
Discuss

The nth term of an A.P., the sum of whose n terms is Sn, is

  • 1] Sn + Sn - 1
  • 2] Sn - Sn - 1
  • 3] Sn + Sn + 1
  • 4] Sn - Sn + 1
Solution
10
Discuss

The common difference of the A.P. \(\frac{1}{3}, \frac{{1 - 3b}}{3} , \frac{{1 - 6b}}{3}\)   . . . . . . is

 

  • 1]

    \(\frac{1}{3}\)

  • 2]

    \( - \frac{1}{3}\)

  • 3]

    -b

  • 4]

    b

Solution
# Quiz