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Standard X

Mathematics

Arithmetic Mean

Question 2In ...

Question

Question 2
In an AP, if a = 3.5, d = 0 and n = 101, then $a_{n}$ will be:

A) 0
B) 3.5
C) 103.5
D) 104.5

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Solution

Given, a = 3.5, d = 0 and n = 101
For an AP,
$a_{n} = a + (n - 1) d$
$\Rightarrow a_{101} = 3.5 + (101 - 1) \times 0$
$\Rightarrow a_{101} = 3.5$

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Q. Question 2
In an AP, if a = 3.5, d = 0 and n = 101, then

a_{n}

will be:

A) 0
B) 3.5
C) 103.5
D) 104.5

Q. In an A.P., if

a = 3.5

d = 0

n = 101

, then

a_{n}

will be

Q. Choose the correct alternative answer for each of the following sub questions.
(1) The sequence –10, –6, –2, 2,...

(A) is an A.P., Reason d= –16	(B) is an A.P., Reason d= 4
(C) is an A.P., Reason d= –4	(D) is not an A.P.

(2) First four terms of an A.P. are ....., whose first term is –2 and common difference is –2.

(A) –2, 0, 2, 4	(B) –2, 4, –8, 16
(C) –2, –4, –6, –8	(D) –2, –4, –8, –16

(3) What is the sum of the first 30 natural numbers ?

(A) 464

(B) 465

(D) 461

(4) For an given A.P. t₇ = 4, d = –4, n = 101, then a = ....

(A) 6

(B) 7

(D) 28

(5) For an given A.P. a = 3.5, d = 0, n = 101, then t_n = ....

(A) 0

(B) 3.5

(D) 14.5

(6) In an A.P. first two terms are –3, 4 then 21^st term is ...

(A) –143

(B) 143

(D) 17

(7) If for any A.P. d = 5 then t₁₈ – t₁₃ = ....

(A) 5

(B) 20

(D) 30

(8) Sum of first five multuiples of 3 is...

(A) 45

(B) 55

(D) 75

(9) 15, 10, 5,... In this A.P sum of first 10 terms is...

(A) –75

(B) –125

(D) 125

(10) In an A.P. 1^st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....

d = - 2, n = 5

and

a_{n} = 0

, then the value of

a

Q. Question 1
In an AP, if d = - 4 , n = 7 and a_n = 4, then a is equal to

A) 6
B) 7
C) 20
D) 28

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Arithmetic Mean

Standard X Mathematics

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Question 2 In an AP, if a = 3.5, d = 0 and n = 101, then an will be: A) 0 B) 3.5 C) 103.5 D) 104.5

Given, a = 3.5, d = 0 and n = 101 For an AP, an=a+(n−1)d ⇒a101=3.5+(101−1)× 0 ⇒a101=3.5

Question 2
In an AP, if a = 3.5, d = 0 and n = 101, then $a_{n}$ will be:

A) 0
B) 3.5
C) 103.5
D) 104.5

Given, a = 3.5, d = 0 and n = 101
For an AP,
$a_{n} = a + (n - 1) d$
$\Rightarrow a_{101} = 3.5 + (101 - 1) \times 0$
$\Rightarrow a_{101} = 3.5$