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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, ....., 2253,9,15,21,.....,225 ?

Solution

ddd === 9−3=69 - 3 = 69−3=6

n=xn−x1d+1=225−36+1=38n=\dfrac {x_n- x_1}{d}+1=\dfrac {225-3}{6}+1 = 38n=dxn​−x1​​+1=6225−3​+1=38

Answer: 383838 terms.

Example 2

The first term of an A.P is 141414 and the last term is 146146146. If the sum of the terms in the A.P is 184018401840, then how many terms are there in this A.P?

Solution

Average of AP === 14+1462=80\dfrac {14 + 146}{2} = 80214+146​=80

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

Answer: 232323

Example 3

What is the sum of terms for the following Arithmetic Progressions:
(I) 37,43,4937, 43, 4937,43,49 ... 355355355
(II)57,53,49 57, 53, 4957,53,49 ... 30terms_{30terms}30terms​

Solution

Case I: d=d =d= 43−37=643 - 37 = 643−37=6.

n=355−376+1=54n = \dfrac {355 - 37}{6} + 1 = 54n=6355−37​+1=54
Sum of the terms === 54×(37+3552)=54×19654 \times \left( \dfrac{37 + 355}{2} \right) = 54 \times 19654×(237+355​)=54×196 = 10584

Case II: a=57a = 57a=57, ddd = 53−57=−453 - 57 = -453−57=−4, n=30n = 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)]2n​×[2a+(n−1)d]=230​×[114+(30−1)×(−4)]

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

Answer: (I) 1058410584 10584    (II) −30-30 −30

Example 4

xnx_nxn​ is the nthn^{\text{th}}nth term of an AP where the sum of x3x_3x3​ and x4x_4x4​ equals the sum of x5x_5x5​,x6x_6x6​, andx7x_7x7​. Which of the following equals 0?

(1)x10x_{10}x10​             (2)x11x_{11}x11​            (3)x12x_{12}x12​            (4)x13x_{13}x13​

Solution

x3+x4=x5+x6+x7x_3 + x_4 = x_5 + x_6 + x_7x3​+x4​=x5​+x6​+x7​
⇒ (a+2d)+(a+3d)(a + 2d) + (a + 3d)(a+2d)+(a+3d) = (a+4d)+(a+5d)+(a+6d)(a + 4d) + (a + 5d) + (a + 6d)(a+4d)+(a+5d)+(a+6d)
⇒ 2a+5d=3a+15d2a + 5d = 3a + 15d2a+5d=3a+15d
⇒ a+10d=0a + 10d = 0a+10d=0
⇒ x11=0x_{11}=0x11​=0

Answer: (2)x11(2) x_{11}(2)x11​

Example 5

What is the sum of all 222 digit numbers divisible by 777?

Solution

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

n=98−147+1=13 n = \dfrac {98-14}{7}+1 = 13n=798−14​+1=13

Sum of terms === 13×(14+982)=13×56=72813 \times \left( \dfrac{14 + 98}{2} \right) = 13 \times 56 = 72813×(214+98​)=13×56=728

Answer: 5 5 5  or  11 1111

Example 6

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

Solution

This can be categorised as 2 APs.

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

⇒ 11×(1+1012)11 \times \left( \dfrac{1 + 101}{2} \right)11×(21+101​) - 10×(6+962)10 \times \left( \dfrac{6 + 96}{2} \right) 10×(26+96​) = 11×51−10×5111 \times 51 - 10 \times 51 11×51−10×51 = 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 = 555

Total terms in this sequence = n=101−15+1=21n = \dfrac {101 - 1}{5} + 1 = 21n=5101−1​+1=21

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

=== 1+(–6+11)+(–16+21)+...+(–96+101)1 + (–6 + 11) + (–16 + 21) + ... + (–96 + 101)1+(–6+11)+(–16+21)+...+(–96+101)
From the last 202020 terms we form 10\bm{10}10 pairs, where the value of each pair is 555.

x=1+5×10=51x = 1 + 5 \times 10 = 51x=1+5×10=51

x=51x = 51x=51

Answer: 5151 51

Example 7

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

Solution

Let aaa and ddd be the first term and common difference.

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

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

⇒ 2a−d+dn=8n+22a - d + dn = 8n + 22a−d+dn=8n+2

Comparing coefficients of nnn, we get dndndn = 8n8n8n ⇒ d=8\bm{d = 8}d=8

Comparing constants we get 2a2a2a – ddd = 222 ⇒ a=5\bm{a = 5}a=5

10th term of AP = a+9da + 9da+9d = 5+9×85 + 9 \times85+9×8 = 77\bm{77}77

Answer: 7777 77

Example 8

If the sum of 5,8,115, 8, 115,8,11, ... and x x x is 220220220, then how many terms are there in this AP?

(1) 888               (2) 999                (3) 101010               (4) 111111

Solution

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

Sn = n2×[2a+(n−1)d]\dfrac{n}{2} \times [2a + (n-1)d]2n​×[2a+(n−1)d] = 220220220

⇒ n2×[2a+(n−1)×3]=220\dfrac{n}{2} \times [2a + (n-1)\times 3] = 2202n​×[2a+(n−1)×3]=220

⇒ n(3n+7)=440n(3n+7) = 440n(3n+7)=440

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

Answer:(4)11(4) 11 (4)11

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