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CAT 2025 Lesson : Progressions - AP Examples

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Example 1

How many terms are in the AP 3,9,15,21,.....,2253, 9, 15, 21, ....., 225 ?

Solution

dd == 9−3=69 - 3 = 6

n=xn−x1d+1=225−36+1=38n=\dfrac {x_n- x_1}{d}+1=\dfrac {225-3}{6}+1 = 38

Answer: 3838 terms.

Example 2

The first term of an A.P is 1414 and the last term is 146146. If the sum of the terms in the A.P is 18401840, then how many terms are there in this A.P?

Solution

Average of AP == 14+1462=80\dfrac {14 + 146}{2} = 80

Sum of terms of AP == n×Averagen \times \text{Average}
⇒ 1840=n×801840 = n \times 80
⇒ n=23n = 23

Answer: 2323

Example 3

What is the sum of terms for the following Arithmetic Progressions:
(I) 37,43,4937, 43, 49 ... 355355
(II)57,53,49 57, 53, 49 ... 30terms_{30terms}

Solution

Case I: d=d = 43−37=643 - 37 = 6.

n=355−376+1=54n = \dfrac {355 - 37}{6} + 1 = 54
Sum of the terms == 54×(37+3552)=54×19654 \times \left( \dfrac{37 + 355}{2} \right) = 54 \times 196 = 10584

Case II: a=57a = 57, dd = 53−57=−453 - 57 = -4, n=30n = 30

Sum of the terms == n2×[2a+(n−1)d]=302×[114+(30−1)×(−4)] \dfrac {n}{2} \times [2a + (n-1)d]= \dfrac {30}{2} \times [114 + (30 - 1)\times (- 4)]

⇒ 15×(114−116)=15 \times (114-116)= -30

Answer: (I) 1058410584    (II) −30-30

Example 4

xnx_n is the nthn^{\text{th}} term of an AP where the sum of x3x_3 and x4x_4 equals the sum of x5x_5,x6x_6, andx7x_7. Which of the following equals 0?

(1)x10x_{10}             (2)x11x_{11}            (3)x12x_{12}            (4)x13x_{13}

Solution

x3+x4=x5+x6+x7x_3 + x_4 = x_5 + x_6 + x_7
⇒ (a+2d)+(a+3d)(a + 2d) + (a + 3d) = (a+4d)+(a+5d)+(a+6d)(a + 4d) + (a + 5d) + (a + 6d)
⇒ 2a+5d=3a+15d2a + 5d = 3a + 15d
⇒ a+10d=0a + 10d = 0
⇒ x11=0x_{11}=0

Answer: (2)x11(2) x_{11}

Example 5

What is the sum of all 22 digit numbers divisible by 77?

Solution

22-digit multiples of 7 are in AP ⇒ 14,21,...,9814, 21, ... , 98

n=98−147+1=13 n = \dfrac {98-14}{7}+1 = 13

Sum of terms == 13×(14+982)=13×56=72813 \times \left( \dfrac{14 + 98}{2} \right) = 13 \times 56 = 728

Answer: 5 5  or  11 11

Example 6

If xx = 1–6+11–16+21–.......+1011 – 6 + 11 – 16 + 21 –....... + 101, thenx= x = ?

Solution

This can be categorised as 2 APs.

xx == 1–6+11–16+21–.......+1011 – 6 + 11 – 16 + 21 –....... + 101 = (1+11+21+...101)–(6+16+26+...+96)(1 + 11 + 21 + ... 101) – (6 + 16 + 26 + ... + 96)

⇒ 11×(1+1012)11 \times \left( \dfrac{1 + 101}{2} \right) - 10×(6+962)10 \times \left( \dfrac{6 + 96}{2} \right) = 11×51−10×5111 \times 51 - 10 \times 51 = 51

Alternatively (Recommended)

Note that the signs of these terms are alternating between positive and negative. Therefore, sum of adjacent terms will be a constant.

Difference in absolute values of the terms = 55

Total terms in this sequence = n=101−15+1=21n = \dfrac {101 - 1}{5} + 1 = 21

xx == 1–6+11–16+21–.......+1011 – 6 + 11 – 16 + 21 –....... + 101

== 1+(–6+11)+(–16+21)+...+(–96+101)1 + (–6 + 11) + (–16 + 21) + ... + (–96 + 101)
From the last 2020 terms we form 10\bm{10} pairs, where the value of each pair is 55.

x=1+5×10=51x = 1 + 5 \times 10 = 51

x=51x = 51

Answer: 5151

Example 7

If sum of n terms of an Arithmetic Progression is n(4n+1)n(4n + 1), then the 10th10^{\text{th}} term of this sequence is

Solution

Let aa and dd be the first term and common difference.

Sn = n2×[2a+(n−1)d]\dfrac{n}{2} \times [2a + (n - 1)d] = n(4n+1)n(4n + 1)

⇒ [2a+(n−1)d]=8n+2[2a + (n - 1)d] = 8n + 2

⇒ 2a−d+dn=8n+22a - d + dn = 8n + 2

Comparing coefficients of nn, we get dndn = 8n8n ⇒ d=8\bm{d = 8}

Comparing constants we get 2a2a – dd = 22 ⇒ a=5\bm{a = 5}

10th term of AP = a+9da + 9d = 5+9×85 + 9 \times8 = 77\bm{77}

Answer: 7777

Example 8

If the sum of 5,8,115, 8, 11, ... and x x is 220220, then how many terms are there in this AP?

(1) 88               (2) 99                (3) 1010               (4) 1111

Solution

a=5a = 5, d=3d = 3 and let xx be the nth n^{\text{th}} term.

Sn = n2×[2a+(n−1)d]\dfrac{n}{2} \times [2a + (n-1)d] = 220220

⇒ n2×[2a+(n−1)×3]=220\dfrac{n}{2} \times [2a + (n-1)\times 3] = 220

⇒ n(3n+7)=440n(3n+7) = 440

Substituting from the options, only n=11\bm{n = 11} satisfies.

Answer:(4)11(4) 11

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