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Which of the following is not an A.P. ?  Option: 1 Option: 2 Option: 3 Option: 4

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

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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

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

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The first term of an  is  and the common difference is , then its  term is  Option:1  Option:2  Option:3  Option:4

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

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The value of  for which  and  are the three consecutive terms of an , is Option: 1 Option: 2 Option: 3 Option: 4

are three consecutive terms. There fore

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.