CAT 2023 Set-3 | Quantitative Aptitude | Question: 20

Question

Let $a_{n}=46+8 n$ and $b_{n}=98+4 n$ be two sequences for natural numbers $n \leq 100$. Then, the sum of all terms common to both the sequences is

• A. $14900$
• B. $15000$
• C. $14798$
• D. $14602$

   

Answer 1

First Let’s talk about each term individually.

b_n = 98 + 4n

b₁ = 98 + 4 *1 = 102.

a_n = 46 + 8n

a₁  = 46 + 8 *1 = 54.

Now a₁ < b_1.

Now, let’s do hit and trial and find the first common term in the both A.Ps.

102 = 46 + 8 n  => n = 7.

So first common term in a_n and b_n  = 102.

Now common difference will be LCM of (8,4) = 8. [ ∵ 8 is common difference of a_n and 4 is the common difference of bn]

New A.P.  we got is let’s call it c_n, 102, 102 + 8, 102 + 8*2 inshort c_n = 102 + 8*n.

So till now we got first term of c_n.

Now let’s find the last term of c_n.

a_{100 =} 846.

b₁₀₀ = 498.

It is obvious that last term of c_n will be less than or equal to last term of b_n.

Let’s check the whether 498 will be the last common term of c_n.

498 = 102 + 8*(n-1)

8 *(n-1) = 396.

n = 49.5.

so 498 will not be the last term,

let's check 492 of bn  so we will get 49.

So 492 is the last term of c_n.

For c_n ; a = 102, d = 8 , last term = 492, total terms = 49.

Using sum formula

S_n = 49 / 2[492 + 102]

S_n = 49 * 298

S_n = 14602

Option D is the correct answer.

Comments

there is a slight mistake 

So 492 is the last term of c_n.

the last term would be $494$ and not $492$

$C_n=102+8k$ , $k=0,1,2,...,49$

$S_n=\frac{50}{2}(204+49*8)=14,900$

Answer 2

Option A is the answer

Reason:

Given,

a_n = 46 + 8n = 6 + 8(5 + n)

b_n = 98 + 4n = 2 + 4(24 + n)

Now understand how a_n and b_n are related

a_n on divides by 8 gives remainder 6 => On divided by 4 gives remainder 2

b_n on divides by 4 also gives remainder 2

Now let's observe how the sequence of a_n and b_n look like

a_n = {54, 62, 70,..........,846}

b_n = {102, 106, 110, 114,..........,498}

The elements in b_n which on divided by 8 gives remainder 2 are subset of a_n

102÷8 = 6 => our desired sequence = {102, 110,..........}

Now let's find last term,

498÷8 = 2 => 494 will be last term in our desired sequence. Because, every alternative term in b_n belongs to a_n

Finally sum of ou sequence = 102 + 110 + ...........494 = S_n

First term = 102, Last term = 494

=> 494 = 102 + (n - 1).8

=> 392/8 = n - 1

=> n = 50

S_n = (50 / 2)(102 + 494) = 25 x 596 = 14900

Therefore, Option A is the answer