If 100 times the 100th term of an AP with non zero common differ

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

Advertisement

151.

If 100 times the 100th term of an AP with non zero common difference equals the 50 times its 50th term, then the 150th term of this AP is

  •  –150

  • 150

  • times its 50th term

  • times its 50th term


D.

times its 50th term

The 150 th term of this AP
Let a be the first term and d be the common difference of the given AP, then
T100 =  a+ (100-1)d = a + 99d
T50 = a +(50-1)d = a +49 d
T150 = a + (150-1) d = a +149 d
Now, according to the question,
100 x T100 = 50 x T50
⇒ 100 (a +99d) = 50(a +49d)
2(a +99d) = (a+ 49d)
2a +198 d =a +49d
a +149d = 0

263 Views

Advertisement
152.

Let x1, x2, ......, xn be n observations, and let top enclose straight x be their arithmetic mean and σ2 be their variance.
Statement 1: Variance of 2x1, 2x2, ......, 2xn is 4 σ2.
Statement 2: Arithmetic mean of 2x1, 2x2, ......, 2xn is 4top enclose straight x.

  • Statement 1 is false, statement 2 is true

  • Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

  • Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

213 Views

153.

A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after

  • 18 Months

  • 19 Months

  • 20 Months

  • 20 Months

140 Views

154.

A person is to count 4500 currency notes. Let a denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ...are in an AP with common difference -2, then the time taken by him to count all notes is

  • 24 min

  • 34 min

  • 125 min

  • 125 min

349 Views

Advertisement
155. Let space straight S subscript 1 space equals space sum from straight j space equals 1 to 10 of space straight j space left parenthesis negative 1 right parenthesis space to the power of 10 straight C subscript straight j space comma space straight S subscript 2 space equals space sum from straight j space equals 1 to 10 of space straight j to the power of 10 straight C subscript straight j space and space straight S subscript 3 space equals sum from straight j space equals 1 to 10 of space straight j squared space to the power of 10 straight C subscript straight j

Statement-1: S3 = 55 × 29.
Statement-2: S1 = 90 × 28 and S2 = 10 × 28.

  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

  • Statement-1 is true, Statement-2 is true; statement-2 is not a correct explanation for Statement-1. 

  • Statement-1 is true, Statement-2 is false.

  • Statement-1 is true, Statement-2 is false.

160 Views

156.

For any three positive real numbers a, b and c, (25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c), Then

  • b , c and a are in G.P

  • b, c and a are in A.P

  • a, b and c are in A.P

  • a, b and c are in A.P

814 Views

157.

Let a, b, c ∈ R. If f(x) = ax2 + bx + c is such
that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, ∀x,y ∈ R,then sum from straight n space equals 1 to 10 of space straight f space left parenthesis straight n right parenthesis is equal to

  • 225

  • 330

  • 165

  • 165

330 Views

158.

The sum to infinity of the series 1 space plus space 2 over 3 space plus space 6 over 3 squared space plus space 10 over 3 cubed space plus space 14 over 3 to the power of 4 space plus.... space is

  • 2

  • 3

  • 4

  • 4

147 Views

Advertisement
159.

Statement 1: The variance of first n even natural numbers is fraction numerator straight n squared minus 1 over denominator 4 end fraction
Statement 2: The sum of first n natural numbers is fraction numerator straight n space left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction and the sum of squares of first n natural numbers is fraction numerator straight n left parenthesis straight n plus 1 right parenthesis left parenthesis 2 straight n plus 2 right parenthesis over denominator 6 end fraction

  • Statement–1 is true, Statement–2 is true, Statement–2 is a correct explanation for statement–1

  • Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1.

  • Statement–1 is true, statement–2 is false.

  • Statement–1 is true, statement–2 is false.

265 Views

160.

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals

  • 1 half left parenthesis 1 minus square root of 5 right parenthesis
  • 1 half square root of 5
  • square root of 5
  • square root of 5
330 Views

Advertisement