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Arithmetic Progression (A.P.)

 A sequence is said to be in Arithmetic Progression when they increase or decrease by a constant number. This constant number is called the common difference (c, d) of the arithmetic progression.

         1, 3, 5, 7, ………......             c.d. = 2

        –7, –3, 1, 5, 9 ………             c.d. = 4

        8, 5, 2, –1, –4 ………             c.d. = –3

Since we are adding ‘d’ (common difference) each time (negative value of d accounts here for subtraction) to get next number in the sequence. So by close inspection we can easily say nth term of an A.P. will be given by

        tn = a + (n–1)d.

 

Sum of the first n term of an A.P.

 Our next interest is to find the sum of first ‘n’ terms of an A.P. Let us denote it by Sn

Sn = {a} + {a + d} + {a + 2d} +…+ {a + (n–1)d}                  … (1)

we can write the above series in reverse way also.

Sn = {a+(n–1)d} + {a+(n–2)d} + {a+(n–3)d} +……+ {a}         … (2)

Adding (1) and (2)

2Sn = {2a(n–1)d} + {2a(n–1)d} + (2a+(n–1)d}+…+ {2a+(n–1)d}

or 2 Sn = n {2a+(n–1)d}

Sn =  {2a + (n–1)d}                                                           … (3)

Once can also remember the above formula, in this way

        Sn = first term+last term/2 × number of terms

Suppose, now you have to find out the sum of j-th term of k-th term of an A.P.

        Sj.k = jth term+kth term/2 × (k – j + 1)

(We are also including here j-th term so number of terms = k – j + 1)

or     Sj.k =  a+(j–1)d+a+(k–1)d/2 (k – j + 1)

or     Sj.k = (k–j+1)/2 {2a + (j – k + 2)d}                                     … (4)

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