Study Notes for CAT 2021: 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 n^th term of an A.P. will be given by

t_n = 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 S_n

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

we can write the above series in reverse way also.

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

Adding (1) and (2)

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

or 2 S_n = n {2a+(n–1)d}

S_n = {2a + (n–1)d} … (3)

Once can also remember the above formula, in this way

S_n = 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.

S_j.k = j^th term+k^th term/2 × (k – j + 1)

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

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

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

Example 1:

Let sum of n terms of a series be n (2n–1). Find its mth terms.

Solution:

Let S_m and S_m–1 denote the sum of first m and (m – 1) terms respectively.

S_m = T₁ + T₂ + T₃ + ……. + T_m–1 + T_m

S_m = T₁ + T₂ + T₃ + ……. + T_m–1

Subtracting

S_m – S_m–1 = T_m

⇒ T_m = (m(2m–1))–(m–1)(2(m–1)–1))

= (2m² – m)–(2m² – 5m + 3)

= 4m – 3

Example 2:

The sums of n terms of two A.P.’s are in the ratio 3n + 2: 2n + 3. Find the ratio of their 10^th terms.

Solution:

Let a, a + d, a + 2d, a + 3d, ……………

A, A + D, A + 2D, A + 3D, ………………

be two A.P.’s

n/2[2a+(n–1)d]/n/2[2A+(n–1)d] = 3n+2/2n+3 (given)

⇒ a+n–1/2d/A+n–1/2D = 3n+2/2n+3

⇒ To get the ratio of 10^th terms put n–1/2 = 9

or n = 19

⇒ a+9d/A+9D = 3(19)+2/2(19)+3 = 59/41

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