Posted on - 07-02-2017

IIT JEE

A sequence (progression) is a set of numbers in a definite order with a definite rule of obtaining the numbers.

An A.P. is a sequence whose terms increase or decrease by a fixed number, called the common difference of the A.P. .

If a is the first
term and d the common difference, the A.P. can be written

as a, a+d, a + 2d, ...... . The nth term a_{n} is given by a_{n}
= a + (n - 1)d. The sum S_{n} of the first n terms of such an A.P. is
given by (a + l ) where l is
the last term (i.e. the nth term of the A.P.).

Notes:

- If a fixed number is added (subtracted) to each term of a given A.P. then the resulting sequence is also an A.P. with the same common difference as that of the given A.P.
- If each term of an A.P. is multiplied by a fixed number(say k) (or divided by a non-zero fixed number), the resulting sequence is also an A.P. with the common difference multiplied by k.
- If a
_{1}, a_{2}, a_{3}.....and b_{1}, b_{2}, b_{3}...are two A.P.’s with common differences d and d¢ respectively then a_{1}+b_{1}, a_{2}+b_{2}, a_{3}+b_{3},...is also an A.P. with common difference d+d¢ . - If we have to take
three terms in an A.P., it is convenient to take them as a - d, a,

a + d. In general, we take a - rd, a - (r - 1)d,......a - d, a, a + d,.......a + rd in case we have to take (2r + 1) terms in an A.P. - If we have to take four terms, we take a - 3d, a - d, a + d, a + 3d. In general, we take a - (2r - 1)d, a - (2r - 3)d,....a - d, a + d,.....a + (2r - 1)d, in case we have to take 2r terms in an A.P.
- If a
_{1}, a_{2}, a_{3}, ……. a_{n}are in A.P. then a_{1}+ a_{n}= a_{2}+ a_{n-1}= a_{3}+ a_{n –2}= . . . . . and so on.

- If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two i.e. if a, b, c are in A.P. then is the A.M. of a and c.
- If a
_{1}, a_{2}, ... a_{n}are n numbers then the arithmetic mean (A) of these numbers is_{} - The n numbers A
_{1}, A_{2}......A_{n}are said to be A.M.’s between the numbers a and b_{}if. - a, A
_{1}, A_{2},........A_{n,}b are in A.P. If d is the common difference of this A.P. then. - b = a + (n + 2 - 1)d

&⇒

&⇒ where A_{r}is the r^{th}mean.

If the I^{st}
and the 2^{nd} terms of an A.P are 1 and –3 respectively, find the n^{th
}term and the sum of the I^{st} n terms.

Ist term = a, 2nd term = a + d where a = 1, a + d = -3,

&⇒ d = – 4 (Common difference of A.P.).

we have a_{n}
= a + (n –1)d

= 1 + (n – 1) (– 4) = 5 – 4n

S_{n} = {a + a_{n}}
= {1 + 5 – 4n} = n (3
– 2n)

If 6 arithmetic means
are inserted between 1 and 9/2, find the 4^{th} arithmetic mean.

Let a_{1},
a_{2}, a_{3}, a_{4}, a_{5}, a_{6} be
six arithmatic means .

Then 1, a_{1},
a_{2}, …, a_{6}, will be in A.P.

Now, = 1 + 7d

&⇒ = 7d

&⇒ d =

Hence a_{4}
= 1 + 4 = 3