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Which of the following is not an A.P. ? 
Option: 1 -1.2, 0.8, 2.8, ...
Option: 2 3, 3 + \sqrt 2, 3 + 2\sqrt 2, 3 + 3 \sqrt2, ...
Option: 3 \frac{4}{3},\frac{7}{3},\frac{9}{3},\frac{12}{3},...
Option: 4 \frac{-1}{5},\frac{-2}{5},\frac{-3}{5},...

In option a the series has the same common difference and option b the series has the same common difference hence both are in A.P..

Check for Option C

\frac{4}{3},\frac{7}{3},\frac{9}{3},\frac{12}{3} \\ \\ a_2 - a_1 = \frac{7}{3} - \frac{4}{3} = 1 \\\\ a_3 - a_2 = \frac{9}{3} - \frac{7}{3} = \frac{2}{3} \\ $ Hence it is not an A.P.

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Safeer PP

The 9th term of the A.P. – 15, – 11, – 7, ...., 49 is
Option: 1 32
Option: 2 0
Option: 3 17
Option: 4 13

\\ a = -15, d = -11-(-15) = 4 , n =9\\ T_9 = a+(9-1)d \\ T_9 = -15 + (9-1) \times 4 \\ T_9 = -15 + 32 \\ T_9 = 17\\ $The 9th term of the A.P is 17.

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Safeer PP

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The first term of an A.P. is 5 and the last term is 45. If the sum of all the terms is 400, the number of terms is
Option: 1 20
Option: 2 8
Option: 3 10
Option: 4 16

First term (a)=  5

last term (l) = 45

\\S_n = \frac{n}{2}(a+l) \\ 400 = \frac{n}{2}(5+45) \\ n = 16

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Safeer PP

The value of p for which (2p + 1), 10 and (5p + 5) are three consecutive terms of an AP is 
Option: 1 -1
Option: 2 -2
Option: 3 1
Option: 4 2

The common difference of an ap will be the same

\\10-(2p+1)=5p+5-10\\9-2p=5p-5\\7p=14\\p=2

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Safeer PP

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The common difference of an AP, whose nth term is an = (3n + 7), is 
Option: 1 3
Option: 2 7
Option: 3 10
Option: 4 6

\\a_n=a+(n-1)d=a-d+nd\\given\ a-d+nd=3n+7\\implies\ d=3\ and \ a-d=7

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Safeer PP

The nth term of the A.P. a, 3a, 5a, …… is
Option: 1 na
Option: 2 (2n-1)a
Option: 3 (2n+1)a
Option: 4 2na

\begin{array}{l} a, 3 a, 5 a .... \\ d = 2a \\ a_{n}=a+(n-1) \cdot 2 a \\ =a[1+2 n-2] \\ =a[2 n-1] \end{array}

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The common difference of the A.P .\dpi{100} \small \frac{1}{p}, \frac{1-p}{p}, \frac{1-2p}{p},. . . . . is  
Option: 1 1
Option: 2 \frac{1}{p}
Option: 3 -1
Option: 4 -\frac{1}{p}

\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}.... \\\\ \frac{1}{p}, \frac{1}{p}-1, \frac{1}{p}-2 ..\\\\ $Common difference $(d)= -\frac{1}{p}-1-\frac{1}{p}=-1

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Safeer PP

The first term of an AP is p and the common difference is q, then its 10^{th} term is 
Option:1  q+9p
Option:2  p-9q
Option:3  p+9q
Option:4  2p+9q

a_n=a+(n-1)d

Here the first term a=p and the comment difference d=q. Therefore the tenth term

a_{10}=p+(10-1)q=p+9q

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The value of x for which 2x,(x+10) and (3x+2) are the three consecutive terms of an AP, is
Option: 1 6
Option: 2 -6
Option: 3 18
Option: 4 -18

2x,\ x+10,\ 3x+2 are three consecutive terms. There fore

\\x+10-2x=3x+2-(x+10)\\10-x=2x-8\\3x=18\\x=\frac{18}{3}=6

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Safeer PP

The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

Let she fixes 13 flags on one side and 13 on other side of the middle flag.

distance covered for fixing first flag and return is 2m + 2m = 4m

distance covered for fixing second flag and return is 4m + 4m = 8m

distance covered for fixing third flag and return is 6m + 6m = 12m

Similarly for thirteen flags is

4m + 8m + 12m + …….

Here, a = 4

d = 8 - 4 = 4

n = 13

\\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ \mathrm{S}_{13}=\frac{13}{2}[2(4)+(13-1) 4] \\ \frac{13}{2}[56]=364

Total distance covered = 364 × 2 = 728 m         (for both side )

distance, only to carry the flag = 364.

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infoexpert21

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