Arithmetic Progression for IIT JEE

An Arithmetic Progression (AP) or an arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms in the entire sequence is always constant. This constant term is called the common difference. Let, a be the first term and d be the common difference of an AP, then, the arithmetic progression can be written as:

a, a + d, a + 2d, a + 3d, a + 4d , . . . . . . . . a + (n – 1) d [Where, n = total number of terms in an AP]

The nth term of an AP can be written as tn=a+(n1)d\mathbf{ t_{n} = a + (n – 1) d }, where d=anan1d = a_{n} – a_{n -1}

The Sum of 1st n terms of an Arithmetic Progression is given by:

SnS_{n} = n2\mathbf{\frac{n}{2}} [2a + (n – 1) d]

Since, the last term of an Arithmetic Progression is, l = a + (n – 1) d. Therefore, SnS_{n} = n2\mathbf{\frac{n}{2}} [a + l]

Properties of Arithmetic Progression (AP)

1.The value of the common difference (d) of an AP is either positive or negative number. The AP will be an increasing arithmetic sequence if the value of the common difference (d) is positive. Similarly, the AP will be a decreasing arithmetic sequence if the value of the ‘common difference’ (d) is positive.

Example: 12, 24, 36, 48, . . . . . . . , 120 is an increasing sequence with initial term (a) = 12, common difference (d) = 12 and an = 120. Similarly, 24, 12, 0, – 12, – 24, . . . . . . – 120 a decreasing sequence with the initial term (a) = 24, common difference (d) = – 12 and an = – 120. In this case, the common difference (D) is negative i.e. (d < 0) hence, a1 > a2 > a3 > a3 > . . . . . . > an.

  1. If each term of an Arithmetic Progression (AP) is decreased, increased, divided or multiplied or by the same non-zero number, then the resulting sequence will also form an AP.

Example: Consider an arithmetic progression 4, 8, 12, 16, . . . . . Here, the initial term, a = 4 and the common difference, d = 4. If we add every term of the given sequence by 4 then the resulting sequence 8, 12, 16, 20, . . . will also form an AP with a = 8 and d = 12. Similarly, if we multiply every term of the given sequence by 2 then the resulting sequence 8, 16, 24, 32, . . . . . will also form an AP with a = 8 and d = 8.

  1. If m, n, and o are 3 consecutive terms of an AP, then, 2n = m + o. Also, if m, n, o, and p are 4 consecutive terms of an arithmetic progression then, m + p = n + o.
  2. In an Arithmetic progression, the sum of any 2 terms equidistant from the beginning and end is always constant. Also, their sum is always equal to the sum of 1st term and the last term. Let, a1,a2,a3,a4,...an1,ana_{1}, a_{2}, a_{3}, a_{4}, . . . a_{n – 1}, a_{n} forms an AP, then,

a2+an2a_{2} + a_{n – 2} = a3+an3=a4+an4=a1a_{3} + a_{n – 3} = a_{4} + a_{n – 4} = a_{1} (First Term) + ana_{n} (Last Term)

  1. Any term of an Arithmetic Progression (excluding the Initial term) is equal to the half the sum of terms which are equidistant from it. Let, a1,a2,a3,a4,...an1,ana_{1}, a_{2}, a_{3}, a_{4}, . . . a_{n – 1}, a_{n} forms an AP, then,

a5a_{5} = 12\mathbf{\frac{1}{2}} [a1+a9a_{1} + a_{9}]= 12\mathbf{\frac{1}{2}} [a1+a9][a_{1} + a_{9}] = 12\mathbf{\frac{1}{2}} [a2+a8][a_{2} + a_{8}] = 12\mathbf{\frac{1}{2}} [a3+a7][a_{3} + a_{7}] and so on.

  1. If the nth term of any sequence is a linear expression in n i.e. Pn + Q, then the sequence is an AP and the common difference of that Arithmetic Progression is P.
  2. If the sum of n terms of any sequence is quadratic in n i.e. Pn2+Qn+R,Pn^{2} + Qn + R, then the sequence is an AP and the common difference of that Arithmetic Progression is 2P.

Middle Term of an Arithmetic Progression:

If there are n number of terms in an AP, then,

Case 1: If n is odd

The Middle Term of an Arithmetic Progression is given by (n  +  12)th\mathbf{\left ( \frac{n\;+\;1}{2} \right )^{th}}

Case 2: If n is Even

Middle Terms of an Arithmetic Progression are given by:

(n2)th\mathbf{\left ( \frac{n}{2} \right )^{th}} and (n2  +  1)th\mathbf{\left ( \frac{n}{2}\;+\;1 \right )^{th}}

Arithmetic Progression Important tips:

  1. The Sum of first n natural numbers is given by:

  n  =  n  (n  +  1)2,  n    N\mathbf{\sum \;n\;=\;\frac{n\;(n\;+\;1)}{2},\;n\;\in \;N}

  1. The Sum of the squares of first n natural numbers is given by:

  n2  =  n  (n  +  1)  (n  +  2)6,  n    N\mathbf{\sum \;n^{2}\;=\;\frac{n\;(n\;+\;1)\;(n\;+\;2)}{6},\;n\;\in \;N}

  1. The Sum of the cubes of first n natural numbers is given by:

  n3  =  [n  (n  +  1)2],  n    N\mathbf{\sum \;n^{3}\;=\;\left [ \frac{n\;(n\;+\;1)}{2} \right ],\;n\;\in \;N}

  1. Three conceive terms of an AP can be taken as: (a – d), a, and (a + d). Similarly four consecutive terms of an AP can be taken as: (a – 2d), (a – d), (a + d), and (a + 2d).
  1. Formation of an Arithmetic Progression:

An Arithmetic Progression can be formed by inserting two or more numbers between any two given numbers. Consider P1,P2,P3,P4,P5,....PnP_{1}, P_{2}, P_{3}, P_{4}, P_{5}, . . . . P_{n} be n numbers between k and m such that the resulting series k,P1,P2,P3,P4,P5,....Pn,mk, P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, . . . . P_{n}, m forms an AP. The resulting Arithmetic Progression will have a total of (n + 2) terms.

Hence, m = a + (n – 1) d = k + [(n + 2) – 1] d = k + (n + 1) d [Since, m is (n + 2)th term]

d=mkn+1d = \frac{m-k}{n+1}

Now, P1=a+d=k+mkn+1 P_{1} = a + d = k + \frac {m − k} {n + 1} [Since, n = 1 (Second Term)]

Similarly P2=a+2d=k+2mkn+1 P_{2} = a + 2d = k + 2 \frac {m − k} {n + 1} [Since, n = 1 (Third Term)]

P3=a+3d=k+3mkn+1 P_{3} = a + 3d = k + 3 \frac {m − k} {n + 1} [Since, n = 1 (Fourth Term)]

P4=a+4d=k+4mkn+1 P_{4} = a + 4d = k + 4 \frac {m − k} {n + 1}  [Since, n = 1 (Fifth Term)]

Therefore, for nth term, Pn=a+nd=k+nmkn+1 P_{n} = a + nd = k + n\frac {m − k} {n + 1}

Using this method n numbers can be inserted between two numbers such that the resulting sequence is an AP.

Example: Insert 4 numbers between 12 and 72 such that the resulting sequence is an Arithmetic Progression.

Solution:

Let P1P_{1}, P2P_{2}, P3P_{3}, and P4P_{4} be the four numbers between 12 and 72 such that the sequence 12, P1P_{1}, P2P_{2}, P3P_{3}, and P4P_{4}, 72 forms an AP.

Here, k (Initial Term) = 12, m (Last term) = 72, n (Total No. of Terms) = 6. Therefore, 72 = 12 + (6 -1) d

i.e. 60 = 5d or d = 12

Hence, P1P_{1}­ = a + d = 12 + 12 = 24, P2P_{2}­ = a + 2d = 12 + 24 = 36, P3P_{3} = a + 3d = 12 + 36 = 48, P4P_{4} = a + 4d = 12 + 48 = 60.

Hence, the resulting sequence is 12, 24, 36, 48, 60, 72.

Arithmetic Progressions IIT JEE Problems:

Example 1: If SnS_{n} = 14  n  (  1)k  (k  +  1)2  k2\mathbf{\sum_{1}^{4\;n}\;(-\;1)^{\frac{k\;(k\;+\;1)}{2}}\;k^{2}}. Check if SnS_{n} can take the following values: 1056 and 1088.

Solution:

Given: If SnS_{n} = 14  n  (  1)k  (k  +  1)2  k2\mathbf{\sum_{1}^{4\;n}\;(-\;1)^{\frac{k\;(k\;+\;1)}{2}}\;k^{2}}

i.e. SnS_{n} = 1222+32+425262+72+8292102+112+122+- 1^{2} – 2^{2} + 3^{2} + 4^{2} – 5^{2} – 6^{2} + 7^{2} + 8^{2} – 9^{2} – 10^{2} + 11^{2} + 12^{2} + . . . . . . up to 4n terms.

Or, Sn=(3212)+(4222)+(7252)+(8262)+(11292)+(122102)+S_{n} = (3^{2} – 1^{2}) + (42 – 2^{2}) + (7^{2} – 5^{2}) + (8^{2} – 6^{2}) + (11^{2} – 9^{2}) + (12^{2} – 10^{2}) + . . . . . up to 4n terms.

Or, Sn=8+12+24+28+40+44+S_{n} = 8 + 12 + 24 + 28 + 40 + 44 + . . . . . . up to 4n terms.

Or, Sn=2(4+6+12+14+20+22+S_{n} = 2 (4 + 6 + 12 + 14 + 20 + 22 + . . . . . up to 2n terms).

Or, SnS_{n} = 2 [(4 + 12 + 20 + . . . up to n terms) (6 + 14 + 22 + . . . up to n terms)].

Since, SnS_{n} = n2  [  2a  +  (n    1)  d  ]\mathbf{\frac{n}{2}\;\left [ \;2a\;+\;\left ( n\;-\;1 \right )\;d\; \right ]}

Therefore, SnS_{n} = 2    n2[2  ×  4  +  (n    1)  8]  +  n2  [2  ×  6  +  (n    1)  8]\mathbf{2\;\left | \;\frac{n}{2} \left [ 2\;\times \;4\;+\;\left ( n\;-\;1 \right )\;8 \right ]\;+\;\frac{n}{2}\;\left [ 2\;\times \;6\;+\;(n\;-\;1)\;8 \right ]\right |}

i.e. SnS_{n} = n [8 + 8 n – 8 + 12 + 8n – 8] = 4n ( 4n + 1)

Now, 1088 can be expressed as 32 × 34 = 4.8 (4.8 + 2)

Hence, the value of Sn cannot be 1088.

And, 1056 can be expressed as 32 × 33 = 4.8 (4.8 + 1)

Hence, the value of Sn can be 1056.

Example 2: Find the value of tenth term of an AP if, 4th term and 54th term of an arithmetic progression are 64 and – 61 respectively.

Solution:

Given, T4T_{4} = 64 = a + (4 – 1) d and T54T_{54} = – 61 = a + (54 – 1) d

i.e. a + 3d = 64 . . . . . . . . . (1)

And, a + 53d = – 61 . . . . . . . . (2)

On Subtracting equation (1) and equation (2) we will get:

50 d = – 125 or d =   52\mathbf{-\;\frac{5}{2}}

On substituting the values of d in equation (1) we will get a = 1432\mathbf{\frac{143}{2}}

Now, T10T_{10} = a + 9d = 1432\mathbf{\frac{143}{2}} + 9  ×    52\mathbf{9\;\times \;-\;\frac{5}{2}}

Therefore, T10T_{10} = 49

Example 3: Find 50th term of a sequence if the sum of n terms is given by Sn=2n2+3nS_{n} = 2n^{2} + 3n.

Solution:

Given: Sn=2n2+3nS_{n} = 2n^{2} + 3n

The nth term of the sequence is given by Tn=SnSn1T_{n} = S_{n} – S_{n – 1}

i.e. Tn=2n2+3n[2(n1)2+3(n1)]=2n2+3n2n22+4n3n+3T_{n} = 2n^{2} + 3n – [ 2 (n -1)^{2} + 3(n – 1) ] = 2n^{2} + 3n – 2n^{2} – 2 + 4n – 3n + 3

Or, Tn=4n+1T_{n} = 4n + 1

Hence, T50T_{50} (50th Term) = 4 (50) + 1 = 201.

Example 4: Find the values of a, b, c, and d if,

k  =  1n  [  m  =  1k  m2  ]  =  an4  +  bn3  +  cn2  +  dn  +  e\mathbf{\sum_{k\;=\;1}^{n}\;\left [ \;\sum_{m\;=\;1}^{k}\; m^{2\;}\right ]\;=\;an^{4}\;+\;bn^{3}\;+\;cn^{2}\;+\;dn\;+\;e}

Solution:

Since, the Sum of the squares of first n natural numbers is given by:

  n2  =  n  (n  +  1)  (n  +  2)6,  n    N\mathbf{\sum \;n^{2}\;=\;\frac{n\;(n\;+\;1)\;(n\;+\;2)}{6},\;n\;\in \;N}

Therefore, k  =  1n  [  m  =  1k  m2  ]  =  k  =  1n  k  (k  +  1)  (2k  +  1)6\mathbf{\sum_{k\;=\;1}^{n}\;\left [ \;\sum_{m\;=\;1}^{k}\; m^{2\;}\right ]\;=\;\sum_{k\;=\;1}^{n}\;\frac{k\;(k\;+\;1)\;(2k\;+\;1)}{6}}

= 16  k  =  1n  2k3  +  3k2  +  k\mathbf{\frac{1}{6}\;\sum_{k\;=\;1}^{n}\;2k^{3}\;+\;3k^{2}\;+\;k}

Since, the Sum of the cubes of first n natural numbers is given by:

  n3  =  [n  (n  +  1)2],  n    N\mathbf{\sum \;n^{3}\;=\;\left [ \frac{n\;(n\;+\;1)}{2} \right ],\;n\;\in \;N}

Therefore, 13  [n  (n  +  1)2]2  +  12  [n  (n  +  1)  (2n  +  1)6]  +  112  n  (n  +  1)\mathbf{\frac{1}{3}\;\left [ \frac{n\;(n\;+\;1)}{2} \right ]^{2}\;+\;\frac{1}{2}\;\left [ \frac{n\;(n\;+\;1)\;(2n\;+\;1)}{6} \right ]\;+\;\frac{1}{12}\;n\;(n\;+\;1)}

= 112  [n4  +  n2  +  2n3]  +  112  [2n3  +  3n2  +  n]  +  n212  +  n12\mathbf{\frac{1}{12}\;\left [n^{4}\;+\;n^{2}\;+\;2n^{3}\right ]\;+\;\frac{1}{12}\;\left [2n^{3}\;+\;3n^{2}\;+\;n\right ]\;+\;\frac{n^{2}}{12}\;+\;\frac{n}{12}}

= 16n4+16n3+n23+512n+16\frac{1}{6}n^{4}+\frac{1}{6}n^{3}+\frac{n^{2}}{3}+\frac{5}{12}n+\frac{1}{6}

= 112  n4  +  13  n3  +  512  n2  +  16  n\mathbf{\frac{1}{12}\;n^{4}\;+\;\frac{1}{3}\;n^{3}\;+\;\frac{5}{12}\;n^{2}\;+\;\frac{1}{6}\;n}

On comparing the coefficients of a, b, c, d, and e from the above equation we get:

a = 112\mathbf{\frac{1}{12}}, b = 13\mathbf{\frac{1}{3}}, c = 512\mathbf{\frac{5}{12}}, d = 16\mathbf{\frac{1}{6}} and e = 0.

Example 5: The pth term of the series (31n)+(32n)+(33n)+.\left( 3-\frac{1}{n} \right)+\left( 3-\frac{2}{n} \right)+\left( 3-\frac{3}{n} \right)+…. will be _____.

Solution:

The given series is (31n)+(32n)+(33n)+.\left( 3-\frac{1}{n} \right)+\left( 3-\frac{2}{n} \right)+\left( 3-\frac{3}{n} \right)+….

Therefore common difference d=(32n)(31n)=1nd=\left( 3-\frac{2}{n} \right)-\left( 3-\frac{1}{n} \right)=-\frac{1}{n} and first term a=(31n)a=\left( 3-\frac{1}{n} \right)

Now, pth term of the series=a+(p1)d=(31n)+(p1)(1n)=31n+1npn=(3pn){{p}^{th}} \text  term \ of \ the \ series =a+(p-1)d \\=\left( 3-\frac{1}{n} \right)+(p-1)\left( -\frac{1}{n} \right)\\=3-\frac{1}{n}+\frac{1}{n}-\frac{p}{n}\\=\left( 3-\frac{p}{n} \right)

Trick: This question can be solved by inspection.

first1n,second2n,third3n,therefore,pthwill bepn \text first -\frac{1}{n}, \text second -\frac{2}{n}, \text third-\frac{3}{n}, \text therefore, {{p}^{th}} \text will \ be -\frac{p}{n}\\

Example 6: Which of the following sequences is an arithmetic sequence?

A)f(n)=an+b;nNB)f(n)=krn;nNC)f(n)=(an+b)krn;nND)f(n)=1a(n+bn);nN\text A) f(n)=an+b;\,n\in N\\ \text B) f(n)=k{{r}^{n}};\,n\in N\\ \text C) f(n)=(an+b)\,k{{r}^{n}};\,n\in N\\ \text D) f(n)=\frac{1}{a\left( n+\frac{b}{n} \right)};\,n\in N\\

Solution: 

Sequence f (n) = an + b; n ∈ N is an A.P.

Putting n = 1, 2, 3, 4, ………., we get the sequence

(a + b), (2a + b), (3a + b),……… which is an A.P. where first term (A) = (a + b) and common difference d = a.

Aliter: As we have mentioned in theory part that nth term of an A.P. is of the form an + b, n ∈ N.

Formation of an Arithmetic Progression:

An Arithmetic Progression can be formed by inserting two or more numbers between any two given numbers. Consider P1,P2,P3,P4,P5,....PnP_{1}, P_{2}, P_{3}, P_{4}, P_{5}, . . . . P_{n} be n numbers between k and m such that the resulting series k, P1,P2,P3,P4,P5,....PnP_{1}, P_{2}, P_{3}, P_{4}, P_{5}, . . . . P_{n}, m forms an AP. The resulting Arithmetic Progression will have a total of (n + 2) terms.

Hence, m = a + (n – 1) d = k + [(n + 2) – 1] d = k + (n + 1) d [Since, m is (n + 2)th term]

i.e. d = m    kn  +  1\mathbf{\frac{m\;-\;k}{n\;+\;1}}

Now, P1P_{1} = a + d = k + m    kn  +  1\mathbf{\frac{m\;-\;k}{n\;+\;1}} [Since, n = 1 (Second Term)]

Similarly P2P_{2} = a + 2 d = k + 2  (m    k)n  +  1\mathbf{\frac{2\;(m\;-\;k)}{n\;+\;1}} [Since, n = 1 (Third Term)]

P3P_{3} = a + 3 d = k + 3  (m    k)n  +  1\mathbf{\frac{3\;(m\;-\;k)}{n\;+\;1}} [Since, n = 1 (Fourth Term)]

P4P_{4} = a + 4 d = k + 4  (m    k)n  +  1\mathbf{\frac{4\;(m\;-\;k)}{n\;+\;1}} [Since, n = 1 (Fifth Term)]

Therefore, for nth term, PnP_{n} = a + nd = k + n  (m    k)n  +  1\mathbf{\frac{n\;(m\;-\;k)}{n\;+\;1}}

Using this method n numbers can be inserted between two numbers such that the resulting sequence is an AP.

Example: Insert 4 numbers between 12 and 72 such that the resulting sequence is an Arithmetic Progression.

Solution:

Let P1,P2,P3,P4P_{1}, P_{2}, P_{3}, P_{4} be the four numbers between 12 and 72 such that the sequence 12, P1,P2,P3,P4P_{1}, P_{2}, P_{3}, P_{4}, 72 forms an AP.

Here, k (Initial Term) = 12, m (Last term) = 72, n (Total No. of Terms) = 6. Therefore, 72 = 12 + (6 -1) d

i.e. 60 = 5d or d = 12

Hence, P1P_{1}­ = a + d = 12 + 12 = 24, P2P_{2}­ = a + 2d = 12 + 24 = 36, P3P_{3} = a + 3d = 12 + 36 = 48, P4P_{4} = a + 4d = 12 + 48 = 60.

Hence, the resulting sequence is 12, 24, 36, 48, 60, 72.

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