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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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