Introduction to Arithmetic Progression
An arithmetic progression is a sequence in which each term except the first differs from its preceding term by a constant.
So, if a, b, c be in A.P., this means
\[b-a=c-b\]
\[\Rightarrow 2b=a+c\]
The constant difference is called the common difference of the A.P.
In an A.P. we usually denote the first term by ‘a’, the common difference by ‘d’ and the nth term by t_{n}.
General Form of an Arithmetic Progression
If a is the first term and d is the common difference of an A.P. then the A.P. is
a, a + d, a + 2d, a + 3d, ……, a + (n – 1)d if it is finite and
a, a + d, a + 2d, a + 3d, …… according as it is infinite.
The nth term of the A.P. is
\[{{t}_{n}}=a+\left( n-1 \right)d\]
Arithmetic Mean
If three quantities are in A.P.; then the middle term is called the arithmetic mean between the other two. Thus if a, b, c are in A.P., then b is called the arithmetic mean between a and c.
\[b-a=c-b\]
\[\Rightarrow 2b=a+c\]
\[\therefore b=\frac{a+c}{2}\]
Sum of n terms of an Arithmetic Progression
Let for the given A.P., first term = a, common difference = d, number of terms = n and last term = l.
If the sum of terms of the A.P. be denoted by S,
S = a + (a + d) + (a + 2d) + (a + 3d) + …… + (l – 3d) + (l – 2d) + (l – d) + l
On writing the terms of the above A.P. in reverse order we get,
S = l + (l – d) + (l – 2d) + (l – 3d) + …… + (a + 3d) + (a + 2d) + (a + d) + a
Now adding the two equations, we get:
2S = (a + l) + (a + d + l – d) + (a + 2d + l – 2d) + (a + 3d + l – 3d) + …… + (l – 3d + a + 3d) + (l – 2d + a + 2d) + (l – d + a + d) + (l + a)
= (a + l) + (a + l) + (a + l) + (a + l) + …… n times
=n (a + l)
\[\Rightarrow S=\frac{n}{2}\left( a+l \right)\]
Since, l = a + (n – 1)d
\[\therefore S=\frac{n}{2}\left\{ a+a+\left( n-1 \right)d \right\}\]
\[\therefore S=\frac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}\]
Three or more terms in A.P.
When the sum of three consecutive terms of an A.P. is given, we take the terms as : (a – d), a and (a + d). When the sum of four consecutive terms of an A.P. is given, we take the terms as: (a – 3d), (a – d), (a + d) and (a + 3d). For consecutive five terms in A.P.: take terms as (a – 2d), (a – d), a, (a + d) and (a + 2d). For consecutive six terms in A.P.: take terms as (a – 5d), (a – 3d), (a – d), (a + d), (a + 3d) and (a + 5d).
Properties of an A.P. Series
It may be noted that the character of an A.P. series is unaltered (the common difference between consecutive terms remaining the same) if each of the terms is – increased by a constant quantity (the common difference will be same as before). Diminished by a constant quantity (the common difference will be same as before). Multiplied by a constant quantity k (common difference also multiplied by k). divided by a constant k (common difference also divided by k).
Example 01 |
Which term of the series 1, 5, 9, 13, …, is 101?
Solution:
Here first term (a) = 1, common difference (d) = 5 – 1 = 4.
Let the nth term (t_{n}) = 101.
\[\Rightarrow {{t}_{n}}=a+\left( n-1 \right)d=101\]
\[\Rightarrow 1+\left( n-1 \right)\left( 4 \right)=101\]
\[\Rightarrow 1+4n-4=101\]
\[\Rightarrow 4n=104\]
\[\therefore n=26\]
Therefore, 26^{th} term is 101.
Example 02 |
Find the value of ‘k’ for which k^{2}– 7k, k^{2 }+ 9 and 6 are in A.P.
Solution:
Given, k^{2}– 7k, k^{2 }+ 9 and 6 are in A.P.
\[\therefore \left( {{k}^{2}}+9 \right)-\left( {{k}^{2}}-7k \right)=6-\left( {{k}^{2}}+9 \right)\]
\[\Rightarrow {{k}^{2}}+9-{{k}^{2}}+7k=6-{{k}^{2}}-9\]
\[\Rightarrow 7k+9=-{{k}^{2}}-3\]
\[\Rightarrow {{k}^{2}}+7k+12=0\]
\[\Rightarrow {{k}^{2}}+4k+3k+12=0\]
\[\Rightarrow \left( k+4 \right)\left( k+3 \right)=0\]
Therefore k = -4 and k = – 3.
Example 03 |
Find the 20^{th} term from the end of the sequence 3, 8, 13, …, 253.
Solution:
The given sequence is an A.P. with common difference (d) = 5 and the first term (a) = 3, let the nth term of the sequence be 253.
\[{{t}_{n}}=a+\left( n-1 \right)d=253\]
\[\Rightarrow 3+\left( n-1 \right)\left( 5 \right)=253\]
\[\Rightarrow 3+5n-5=253\]
\[\Rightarrow 5n=255\]
\[\therefore n=51\]
Now, the 20^{th} term from end = (51 – 20 + 1)th term from the beginning
= 32^{nd} term from the beginning
= 3 + (32 – 1) 5 = 3 + (31) 5 = 158.
Example 04 |
If m times the mth term of an A.P. is n times the nth term, then show that the (m + n)th term is a zero.
Solution:
Let a be the first term and d be the common difference of the given A.P.
\[\therefore m{{t}_{m}}=n{{t}_{n}}\]
\[\Rightarrow m\left[ a+\left( m-1 \right)d \right]=n\left[ a+\left( n-1 \right)d \right]\]
\[\Rightarrow am+\left( {{m}^{2}}-m \right)d=an+\left( {{n}^{2}}-n \right)d\]
\[\Rightarrow a\left( m-n \right)+\left\{ \left( {{m}^{2}}-{{n}^{2}} \right)-\left( m-n \right) \right\}d=0\]
\[\Rightarrow \left( m-n \right)\left\{ a+\left( m+n-1 \right)d \right\}=0\]
\[\Rightarrow a+\left( m+n-1 \right)d=0~~\left[ Since,~~m\ne n\Rightarrow m-n\ne 0 \right]\]
\[\therefore {{t}_{m+n}}=0~~\left[ \text{Proved} \right]\]
Example 05 |
Find the sum of all the multiples of 13 between 750 and 1000.
Solution:
The multiples of 13 will form an A.P. whose common difference is 13. Evidently the multiples of 13 next to 750 will be the 1^{st} term and the multiple of 13 just preceding 1000 will be the last term of the series.
Therefore, first term = t_{1} = 750 + 4 = 754 and last term = t_{n} = 1000 – 12 = 988.
Now,
\[{{t}_{n}}=a+\left( n-1 \right)d=988\]
\[\Rightarrow 754+\left( n-1 \right)\left( 13 \right)=988\]
\[\Rightarrow 754+13n-13=988\]
\[\Rightarrow 13n=988-754+13\]
\[\Rightarrow 13n=247\]
\[\therefore n=19\]
Therefore, the required sum
\[{{S}_{19}}=\frac{19}{2}\left( 754+988 \right)=\frac{19}{2}\times 1742=16549\]
Example 06 |
The sum of three numbers in A.P. is 18 and their product is 192. Find the numbers.
Solution:
Let the three numbers in A.P. be a – d, a, a + d.
\[\therefore a-d+a+a+d=18\]
\[\Rightarrow 3a=18\Rightarrow a=6\]
And
\[\left( a-d \right)a\left( a+d \right)=192\]
\[\Rightarrow a\left( {{a}^{2}}-{{d}^{2}} \right)=192\]
\[\Rightarrow 6\left( {{6}^{2}}-{{d}^{2}} \right)=192\]
\[\Rightarrow 36-{{d}^{2}}=\frac{192}{6}=32\]
\[\Rightarrow {{d}^{2}}=36-32\]
\[\Rightarrow {{d}^{2}}=4\Rightarrow d=\pm 2\]
When a = 6, d = 2 the numbers are 4, 6, 8 and when a = 6, d = 2 the numbers are 8, 6, 4.
Therefore required numbers are 4, 6, 8 or 8, 6, 4.
Example 07 |
Divide 24 into 4 parts in A.P. such that the sum of their squares is 164.
Solution:
Let the A.P. be a – 3d, a – d, a + d, a + 3d
\[\therefore a-3d+a-d+a+d+a+3d=24\]
\[\Rightarrow 4d=24\Rightarrow a=6\]
Again,
\[{{\left( a-3d \right)}^{2}}+{{\left( a-d \right)}^{2}}+{{\left( a+d \right)}^{2}}+{{\left( a+3d \right)}^{2}}=164\]
\[\Rightarrow \left( {{a}^{2}}-6ad+9{{d}^{2}} \right)+\left( {{a}^{2}}-2ad+{{d}^{2}} \right)+\left( {{a}^{2}}+2ad+{{d}^{2}} \right)+\left( {{a}^{2}}-6ad+9{{d}^{2}} \right)=164\]
\[\Rightarrow 4{{a}^{2}}+20{{d}^{2}}=164\]
\[\Rightarrow {{4.6}^{2}}+20{{d}^{2}}=164\]
\[\Rightarrow 20{{d}^{2}}=164-144\]
\[\Rightarrow 20{{d}^{2}}=20\]
\[\Rightarrow {{d}^{2}}=1\Rightarrow d=\pm 1\]
Hence the four parts are 6 – 3(±1), 6 – (±1), 6 + (±1) and 6 + 3(±1) i.e., 3, 5, 7, 9 or 9, 7, 5, 3.
Example 08 |
The sums of first p, q, r terms of an A.P. are a, b, c respectively, prove that
\[\frac{a}{p}\left( q-r \right)+\frac{b}{q}\left( r-p \right)+\frac{c}{r}\left( p-q \right)=0\]
Solution:
Let A be the first term and D the common difference. Given S_{p} = a.
\[\therefore \frac{p}{2}\left[ 2A+\left( p-1 \right)D \right]=a\]
\[\Rightarrow A+\frac{p-1}{2}D=\frac{a}{p}……….(i)\]
Again S_{q} = b and S_{r} = c.
\[\Rightarrow A+\frac{q-1}{2}D=\frac{b}{q}……….(ii)~~and~~A+\frac{r-1}{2}D=\frac{c}{r}……….(iii)\]
Multiplying (i) by (q – r), (ii) by (r – p), (iii) by (p – q), and adding, we have
\[A\left( q-r+r-p+p-q \right)+\frac{D}{2}\left\{ \left( p-1 \right)\left( q-r \right)+\left( q-1 \right)\left( r-p \right)++\left( r-1 \right)\left( p-q \right) \right\}\]
\[=\frac{a}{p}\left( q-r \right)+\frac{b}{q}\left( r-p \right)+\frac{c}{r}\left( p-q \right)\]
\[\Rightarrow A\left( 0 \right)+\frac{D}{2}\left( 0 \right)=\frac{a}{p}\left( q-r \right)+\frac{b}{q}\left( r-p \right)+\frac{c}{r}\left( p-q \right)\]
\[\therefore \frac{a}{p}\left( q-r \right)+\frac{b}{q}\left( r-p \right)+\frac{c}{r}\left( p-q \right)=0\]
Example 09 |
In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the First term, Common difference, Sum of the first 20 terms.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the A.P.
Here,
\[{{a}_{4}}=8\Rightarrow a+3d=8……….(i)\]
\[{{a}_{6}}=14\Rightarrow a+5d=14……….(ii)\]
Subtracting (i) from (ii) we have
\[\left( a+5d \right)-\left( a+3d \right)=14-8\]
\[\Rightarrow 2d=6\Rightarrow d=3\]
Putting the value of ‘d’ in (i), we have
\[a+3\left( 3 \right)=8\Rightarrow a=8-9=-1\]
Now the sum of ‘n’ terms
\[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]
\[\therefore {{S}_{20}}=\frac{20}{2}\left[ 2\left( -1 \right)+\left( 20-1 \right)\left( 3 \right) \right]\]
\[\Rightarrow {{S}_{20}}=10\left[ -2+57 \right]\]
\[\therefore {{S}_{20}}=550\] Hence, first term is -1, common difference is 3 and sum of the first 20 terms is 550.
Linear Equations |
Geometric Progression |