Question

Justify whether it is true to say that the following are the nth terms of an AP.
(i) 2n–3 (ii) 3n2+5 (iii) 1+n+n2

NCERT Class X
Mathematics - Exemplar Problems

Chapter _ARITHMETIC PROGRESSIONS

Answer 1

it is an ap because the common difference is 4

Answer 2

Answer:

i) True

ii) False

iii) False

Step-by-step explanation:

/* Arithmetic Progression:

A sequence of numbers such that the difference between the consecutive terms is constant */

$i) Let \:a_{n}= 2n-3$

$if \: n = 1 , \\a_{1} = 2\times 1 - 3 = 2-3\\=-1$

$if \: n = 2 , \\a_{2} = 2\times 2- 3 = 4-3\\=1$

$if \: n = 3 , \\a_{3} = 2\times 3 - 3 = 6-3\\=3$

$a_{2}-a_{1}=1-(-1)=1+1=2$

$a_{3}-a_{2}=3-1=2$

Therefore,

$a_{2}-a_{1}=a_{3}-a_{2}= 2$

Given 2n-3 is nth term of an A.P

$ii) Let \:a_{n}= 3n^{2}+5$

$if \: n = 1 , \\a_{1} = 3(1)^{2}+5\\=3+5=8$

$if \: n = 2 , \\a_{2} = 3(2)^{2}+5\\=12+5\\=17$

$if \: n = 3 , \\a_{3} = 3(3)^{2}+5\\=27+5=32$

$a_{2}-a_{1}=17-8=9$

$a_{3}-a_{2}=32-17=15$

Therefore,

$a_{2}-a_{1}≠a_{3}-a_{2}$

Given 3n²+5 is not nth term of an A.P

$iii) Let \:a_{n}= 1+m+n^{2}$

$if \: n = 1 , \\a_{1} = 1+1+1^{2}\\=3$

$if \: n = 2 , \\a_{2} = 1+2+2^{2}\\=7$

$if \: n = 3 , \\a_{3} = 1+3+3^{2}\\=13$

$a_{2}-a_{1}=7-3=4$

$a_{3}-a_{2}=13-7=6$

Therefore,

$a_{2}-a_{1}≠a_{3}-a_{2}$

Given 1+n+n² is not nth term of an A.P

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