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Progression

August 17, 2020
Category: Progression

☼ Arithmetic Progression ( A.P.) : The progression of the form a, a + d, a + 2d, a + 3d, . . . is known as an A.P. ( Arithmetic Progression or Arithmetic Sequence ) with first term = a and common difference = d.

In an A.P. a, a + d, a + 2d, a + 3d, . . . , we have :

(i) n^th term, T_n = a + ( n – 1 ) d

e.g. (1) Find the 7^th term of 3, 7, 11, . . .

Now, = a + ( n – 1 ) d

= 3 + ( 7 – 1 ) ( 4 )

= 3 + ( 6 ) ( 4 )

= 3 + 24

= 27

(2) Find the 12^th term of 5, 8, 11, 14, . . .

( Try yourself ) Ans. : $T_{12}$ = 38

(ii) Sum of n^th term,

$S_{n} = \frac{n}{2}\left [ 2a +\left (n-1 \right )d \right ]$ ; d = Common difference

= $\frac{n}{2}$ ( a + l ) ; l = $T_{n}$ = Last Term

e.g. (1) Find the sum of first 8 terms of 3, 7, 11, . . .

Now, $S_{n}$ = $\frac{n}{2}$ [ 2a + (n – 1) d ]

$\therefore S_{8}$ = $\frac{8}{2}$ [ 2 (3) + ( 8 – 1) ( 4 ) ]

= 4 [ 6 + ( 7 ) ( 4 ) ]

= 4 [ 6 + 28 ]

= 4 ( 34 )

= 136

(2) Find the sum of first 11 terms of 5, 7, 9, . . .

( Try yourself ) Ans. : $S_{11}$ = 165

(3) Find 9 + 13 + 17 + . . . + 105

Here, a = 9, d = 4 and l = 105

Now, for 9, 13, 17, . . . , 105

Now, $S_{n}$

$\therefore S_{25}=\frac{n}{2} \left (a+l \right )$

$= \frac{25}{2}\left ( 9+105 \right )$

$= \frac{25}{2} \left (114 \right )$

= 25 ( 57 )

= 1425

(4) Find 7 + 10 + 13 + . . . + 64

( Try yourself ) Ans. : $S_{20}$ = 710

(iii) If a, b, c, d, e, f, . . are in A. P., then

b – a = c – b = d – c = e – d = f – e = Common Difference ( d )

(iv) If a, b, c are in A. P., then b is called the arithmetic mean ( A.M.) between a and c. In this case, b =

e.g. (1) Find the A. M. of 3 and 12.

Now, A. M. = $\frac{a+c}{2}$

$A=\frac{3+12}{2}$

= 7.5

(2) Find the A. M. of 10 and 50.

( Try yourself ) Ans. : 30

(v) Arithmetic mean of $a_{1}$ , $a_{2}$ , . . . , $a_{n}$ is = A.M. = $\frac{a_{1}+a_{2}+...+a_{n}}{n}$

e.g. (1) Find the A. M. of 5, 7, 11, 18 and 34.

Now, A.M. = $\frac{a_{1}+a_{2}+...+a_{n}}{n}$

$A=\frac{5+7+11+18+34}{5}$

$=\frac{75}{5}$

$=15$

(2) Find the A. M. of 18, 37, 42, 59.

( Try yourself ) Ans. : 39

(vi) If A, $a_{1}$ , $a_{2}$ , . . . , $a_{n}$ , B are in A.P., then we can say that $a_{1}$ , $a_{2}$ , . . . , $a_{n}$ are the n arithmetic means between A and B.

In this case, d = $\frac{b-1}{n+1}$ and so, = A + d

= A + 2d

. .

= A + nd

Example.

(1) Find five arithmetic means between 7 and 37. or

Fill the blank 7, __, __, __, __, __, 37 with appropriate nos. such that the sequence becomes Arithmetic Sequence.

Now, d = $\frac{b-1}{n+1}$ : Where, a = 7, b = 37 & n = 5

$\therefore d=\frac{37-7}{5+1}$

$= \frac{30}{6}$

$=5$

Nos. are 7 + 5 = 12, 12 + 5 = 17, 17 + 5 = 22, 22 + 5 = 27, 27 + 5 = 32.

i.e. the blanks are 12, 17, 22, 27, 32.

(2) Find eight arithmetic means between 3 and 21.

( Try yourself ) Ans. : 5, 7, 9, 11, 13, 15, 17, 19

(vii) It is convenient to take :

For odd number of terms :

Three terms in A.P. are taken as ( a – d ), a, ( a + d ).

Five terms in A.P. are taken as ( a – 2d ), ( a – d ), a, ( a + d ), ( a + 2d ).

Seven terms in A.P. are taken as

( a – 3d ), ( a – 2d ), ( a – d ), a, ( a + d ), ( a + 2d ), ( a + 3d ).

For even number of terms :

Two terms in A.P. are taken as ( a – d ), ( a + d )

Four terms in A.P. are taken as ( a – 3d ), ( a – d ), ( a + d ), ( a + 3d ).

Six terms in A.P. are taken as

( a – 5d ), ( a – 3d ), ( a – d ), ( a + d ), ( a + 3d ), ( a + 5d ).

e.g. (1) If the sum of 3 terms of an arithmetic sequence is 27 and product is 585, then find those terms.

Let, the terms are ( a – d ), a, ( a + d )

Now,

a – d + a + a + d = 27 & ( a – d ) a ( a + d ) = 585

$\therefore$ 3 a = 27 $\therefore$ ( 9 – d ) 9 ( 9 + d ) = 585

$\therefore$ a = 9 $\therefore$ ( 9 – d ) ( 9 + d ) = 65

$\therefore \left (9 \right )^{2}-\left (d \right )^{2} =65$

$\therefore 81 -\left (d \right )^{2} =65$

$\therefore 81-65 =d ^{2}$

$\therefore d^{2}=16$

$\therefore d=4$

If d = 4, terms are 5, 9, 13. [ using ( a – d ), a, ( a + d ) ]

& If d = – 4, terms are 13, 9, 5. [ using ( a – d ), a, ( a + d ) ]

(2) If the sum of 3 terms of an arithmetic sequence is 33 and product is 935, then find those terms.

( Try yourself ) Ans. : 5, 11, 17 or 17, 11, 5.

Or solve by options directly. It is more easy method.

(3) If the sum of 4 terms of an arithmetic sequence is 20 and product is 384, then find those terms.

$a-3d+a-d+a+d+a+3d=20$ $\left (5-3d \right ) \times \left (5-d \right )\times\left (5+d \right )\times\left (5+3d \right )=384$

$\therefore4a=20$ $\therefore \left (25 - 9d^{2} \right ) \times \left (25 - d^{2} \right )= 384$

$\therefore a=5$ $\therefore 625-25d^{2}-225d^{2}+9d^{4}=384$

$\therefore 9d^{4}-250d^{2}+241=0$

$\therefore \left (d^{2}-1 \right ) \left (9d^{2}-241 \right )=0$

$\therefore d^{2} = 1$ or $d^{2}= \frac{241}{9}$

$\therefore d= \pm1$ or $d= \pm \sqrt{\frac{241}{9}}$ is not possible

If d = 1, then nos. are a – 3d = 2, a – d = 4, a + d = 6, a + 3d = 8

or if d = – 1, then nos. are 8, 6, 4, 2.

(4) If the sum of 4 terms of an arithmetic sequence is 22 and product is 280, then find those terms.

( Try yourself ) Ans. : 1, 3, 7, 10 or 10, 7, 3, 1.

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