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NVAMediumJEE 2023Geometric Progression (GP)

JEE Mathematics 2023 Question with Solution

Let {ak}\{a_k\} and {bk},k∈N\{b_k\}, k \in \mathbb{N}, be two G.P.s with common ratios r1r_1 and r2r_2, respectively, such that a1=b1=4a_1 = b_1 = 4 and r1<r2r_1 < r_2. Let ck=ak+bk,k∈Nc_k = a_k + b_k, k \in \mathbb{N}. If c2=5c_2 = 5 and c3=13c_3 = 13, then Σk=14ck−(12a6+8b4)\Sigma_{k=1}^4 c_k - (12a_6 + 8b_4) is equal to:

Answer

Correct answer:9

Step-by-step solution

Standard Method

Given: ck=ak+bkc_k = a_k + b_k, a1=b1=4a_1 = b_1 = 4, c2=5c_2 = 5 and the solution working gives c3=134c_3 = \frac{13}{4}.

Find: ∑k=1∞ck−(12a6+8b4)\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4) as used in the solution.

For the two G.P.s,

a2=4r1,a3=4r12,b2=4r2,b3=4r22a_2 = 4r_1, \quad a_3 = 4r_1^2, \quad b_2 = 4r_2, \quad b_3 = 4r_2^2

So,

c2=a2+b2=4r1+4r2=5c_2 = a_2 + b_2 = 4r_1 + 4r_2 = 5

Hence,

r1+r2=54r_1 + r_2 = \frac{5}{4}

Also, from the solution,

c3=a3+b3=4r12+4r22=134c_3 = a_3 + b_3 = 4r_1^2 + 4r_2^2 = \frac{13}{4}

Therefore,

r12+r22=1316r_1^2 + r_2^2 = \frac{13}{16}

Using

r12+r22=(r1+r2)2−2r1r2r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2

we get

1316=(54)2−2r1r2\frac{13}{16} = \left(\frac{5}{4}\right)^2 - 2r_1r_2 1316=2516−2r1r2\frac{13}{16} = \frac{25}{16} - 2r_1r_2 2r1r2=1216=342r_1r_2 = \frac{12}{16} = \frac{3}{4} r1r2=38r_1r_2 = \frac{3}{8}

So the common ratios are

r1=12,r2=34r_1 = \frac{1}{2}, \quad r_2 = \frac{3}{4}

because r1<r2r_1 < r_2.

Now,

∑k=1∞ck=∑k=1∞ak+∑k=1∞bk=41−r1+41−r2\sum_{k=1}^{\infty} c_k = \sum_{k=1}^{\infty} a_k + \sum_{k=1}^{\infty} b_k = \frac{4}{1-r_1} + \frac{4}{1-r_2}

Substituting,

∑k=1∞ck=41−12+41−34=8+16=24\sum_{k=1}^{\infty} c_k = \frac{4}{1-\frac{1}{2}} + \frac{4}{1-\frac{3}{4}} = 8 + 16 = 24

Next,

a6=4(12)5=18a_6 = 4\left(\frac{1}{2}\right)^5 = \frac{1}{8}

so

12a6=12⋅18=3212a_6 = 12 \cdot \frac{1}{8} = \frac{3}{2}

And,

b4=4(34)3=2716b_4 = 4\left(\frac{3}{4}\right)^3 = \frac{27}{16}

so

8b4=8⋅2716=2728b_4 = 8 \cdot \frac{27}{16} = \frac{27}{2}

Thus,

12a6+8b4=32+272=1512a_6 + 8b_4 = \frac{3}{2} + \frac{27}{2} = 15

Finally,

∑k=1∞ck−(12a6+8b4)=24−15=9\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4) = 24 - 15 = 9

Therefore, the answer is 99.

Note: The source question text shows Σk=14ck\Sigma_{k=1}^4 c_k and c3=13c_3 = 13, but the solution works with ∑k=1∞ck\sum_{k=1}^{\infty} c_k and c3=134c_3 = \frac{13}{4}.

Common mistakes

  • Using the question text values without reconciling them with the solution. Here the provided working uses c3=134c_3 = \frac{13}{4} and an infinite sum, not c3=13c_3 = 13 and a finite sum to 44. Always follow the authoritative solution source when extracting the final answer.

  • Writing a6=4r16a_6 = 4r_1^6 or b4=4r24b_4 = 4r_2^4. In a G.P. with first term 44, the nnth term is 4rn−14r^{n-1}, so a6=4r15a_6 = 4r_1^5 and b4=4r23b_4 = 4r_2^3.

  • Finding r1+r2r_1 + r_2 and r12+r22r_1^2 + r_2^2 but not using the identity r12+r22=(r1+r2)2−2r1r2r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2. Without this step, the product r1r2r_1r_2 and hence the individual ratios cannot be obtained correctly.

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